## Dual Nature Of Matter And Radiation Quantum Theory Introduction

Quantum theory or quantum physics Is the mainstay of modern physics. This theory is primarily applicable to the microscopic world, i.e., the physics of the atomic domain. The study of tills brunch started almost at the beginning of the twentieth century.

In quantum theory, we come across those phenomena or facts that are beyond our common experience and are non-realistic. Scientists who established the main foundation of this theory were also surprised to see the inferences obtained horn theoretical analysis of the theory. However, nil experiments conducted so far have strengthened the base of the theory.

**Quantized quantity and quantum**

Some quantities, obtained in daily life, can have only chosen values. These values are obtained, generally, by multiplying a primary value by an integer. Such quandaries are called quantized qiiantldcs and the primary value is called a quantum of the respective quantity.

As, the currency is quantized and previously, in Indian currency, 1 paisa was its quantum. It was possible to pay 1 rupee 6 paise or 106 paise but payment of 106.5 paise was not possible.

**Read and Learn More Class 12 Physics Notes**

Scientist Max Planck propounded his quantum theory in 1900 AD. In spite of the astounding success of the wave theory of light, this theory cannot explain phenomena like black body radiations, photoelectric effect, atomic spectra, etc. To explain black body radiation spectrum, Max Planck introduced quantum theory. Later, the concept of photon particles, introduced by Einstein, established the theory more firmly.

**Properties of an electron**

**1. Charge:** Electron is negatively charged. The magnitude of charge of an electron is,

e = 1.6 ×10^{-19 }C in SI

= 4.8 × 10^{-10} esu of charge in the CGS system

From different experiments, we learned that the charge of a body is a quantized quantity and the quantum of charge is the charge of an electron (e). So values of charges can be +2e, -5e, 1000, etc. but values like 1.5e, -2.be are nonrealistic.

**2. Rest mass**: Rest mass of an electron

m_{0} = 9.1 × 10^{-31 }kg = 9.1 × 10^{-28 }g

If the speed of an electron In much lens than the speed of light, there In no difference between its rest mass (m) and effective mass (m). Thus men of electron, m = m_{0}

**3. Kinetic energy of an electron electronvolt:**

The velocity of electron Incrcane on being attracted by a positive potential and hence Its kinetic energy also Increases, On the other hand, when repelled by a negative potential, the velocity of the electron decreases. The kinetic energy of an electron Is usually expressed In the electronvolt (eV) unit.

**Electronvolt Definition:**

The change In kinetic energy of an unbound electron, as it travels across a potential difference of IV, is called 1eV.

1eV= charge of an electron × IV

= 1.6 × 10^{-19}C × IV = 1.6 × 10^{-19} J

∴ C.V = J

= 1.6 × 10^{-19}×10^{7 }erg = 1.6 × 10^{-12} erg

Electronvolt Is a very small unit compared to erg or joule. Hence, It is mainly used In nuclear or atomic physics only.

1keV = 10^{3}eV; 1 MeV = 10^{6}eV

**Quarks**

The discovery of quarks by **Gell-Mann** In 1964 has destroyed the myth that nucleons (protons and neutrons) are the fundamental particles of matter that are incapable of further division and that the charge on the electron was the smallest possible, charge existing In nature.

Quarks have been identified as the fundamental charged particles constituting baryons and mesons. So far, six quarks with their corresponding antiquarks \((\bar{u} \bar{d} \bar{c} \bar{s} \bar{t} \bar{b}\)) have been detected: up (u), down (d), charm (c), strange(s), top (f), bottom (b), with electric charge +\(\frac{2}{3}\)e, and – \(\frac{1}{3}\) e [to be taken alternately in that order], e being the electronic charge.

Thus, for example, a proton is composed of u, u, d, while a neutron is composed of u, d, d, held by mediator particles called gluons. Mesons are composed of quark-antiquark pairs. Incidentally, in the current view.

**All matter consists of three kinds of particles:** Leptons, quarks, and mediators, (details are beyond the scope of the present discussion).

## Dual Nature Of Matter And Radiation Quantum Theory Numerical Examples

**1. What is the energy of a photoelectron, in electronvolt, moving with a velocity of 2 × 10 ^{7} m .s^{-1}? (Given, mass of electron = 9.1 × 10^{-28} g )**

**Solution:**

Mass of electron = 9.1 × 10^{-28} g = 9.1 × 10^{-31} kg

The kinetic energy of the electron

= ½mv² = ½ × ( 9.1 × 10^{-31})×( 2 × 10^{7})²J

= ½ × \(\left\{\frac{9.1 \times 10^{-31} \times\left(2 \times 10^7\right)^2}{1.6 \times 10^{-19}}\right\}\)eV

= 1137.5 eV

## Dual Nature Of Matter And Radiation Photoelectric Effect

**Photoelectric emission Definition:**

The emission of electrons from matter (metals and non-metallic solids) as a consequence of the absorption of energy from electromagnetic radiation of very short wavelengths (such as visible and ultraviolet radiation), is called photoelectric emission.

**Observation of Hertz and Contemporary Scientists**

In 1887 German scientist Hertz observed that when ultraviolet rays fell on the negative electrode of a discharge tube, electric discharge occurred easily.

Subsequently in Hallwach’s experiment two zinc plates were placed in an evacuated quartz bulb and when ultraviolet rays fell on the plate connected to the negative terminal of the battery, immediately a current was found to flow in the circuit. But when the ultraviolet rays fell on the positive plate, there was no flow of current. He also noticed that, as soon as the ultraviolet rays were stopped, the current also stopped. Hallwachs however could not explain this phenomenon

In 1900 Lenard proved that when ultraviolet rays fell on a metallic plate, electrons were emitted from the plate and current was constituted due to the flow of electrons. Since in this case, the flow of current is due to light, it is called the photoelectric effect (photo = light). For this phenomenon, light of short wavelength or high frequency is more effective than light of long wavelength or low frequency. Alkali metals.

**For example:** Lithium, Sodium, Potassium, etc., exhibit a photoelectric effect even in ordinary visible light

**Lenard’s experiment**

G is an evacuated glass bulb with a quartz window Q on its lower face C is a metal plate, kept at potential -V. Another plate A, having a hole at its center is kept at zero potential by earthing. Hence the potential difference between anode and cathode = 0- (-V) = V.

Now, the cathode is illuminated by a monochromatic beam of light, entering through the window. Here ultraviolet rays or visible light of small wavelengths are used according to the nature of cathode plate metal.

Suppose, due to the incidence of light, cathode C emits a beam of negatively charged particles having charge -q. These charged particles are attracted towards the positive plate A. So, the kinetic energy of each charged particle, just before reaching the plate is,

⇒ \(\frac{1}{2} m v^2=q V \quad \text { or, } \frac{q}{m}=\frac{v^2}{2 V}\) ………………………………… (1)

Where, m = mass of each charged particle and v = velocity of the charged particle

A beam of these particles, passing through the hole of the anode is incident on the plate P, which is connected to an electrometer to detect the current. Now, between A and P, a magnetic field B is applied perpendicularly upward concerning the plane of the paper.

Due to this field, the charged particles are forced to move in circular paths. By controlling the magnetic field B, the particles are made incident on the plate D where they follow a circular path of radius of curvature R. The electrometer, connected with the plate D shows the current. Here, the magnetic force, acting on each particle

⇒ \(\vec{F}=-q \vec{v} \times \vec{B}\)

As \(\vec{v} \text { and } \vec{B}\) both are perpendicularÿ to each other, the magnitude of this force,

F = \(\vec{v}\) = qvBsin 90°

= qvB

This force acts, as a centripetal force for the revolving particle.

⇒ \(q v B=\frac{m v^2}{R} \quad \text { or, } \frac{q}{m}=\frac{v}{B R}\)

Or, \(\left(\frac{q}{m}\right)^2=\frac{v^2}{B^2 R^2}\)………………………………….(2)

Dividing the equation (2) by equation (1) we get,

⇒ \(\frac{q}{m}=\frac{2 V}{B^2 R^2}\) ……………………………………. (3)

The value of \(\) can be evaluated by putting the values of V, R, and B in equation (3). From the experimental result thus obtained, Lenard, had shown that the value of \(\frac{q}{m}\) is the same as the specific charge of the electron,\(\frac{e}{m}\) (= 1.76 × 10^{11}C . kg^{-1}) previously known. From this result, it could be concluded that the emitted negatively charged particles from the cathode are electrons

Electrons emitted in this manner, are called photoelectrons. With proper arrangements, the motion of photoelectrons can be made unidirectional. The stream of unidirectional photoelectrons thus produced, develops a current, namely, photoelectric current

**Work function**

The minimum energy required to remove an electron from the surface of a particular substance to a point just outside the surface is called the work function of that substance.

Here the final position of an electron is far from the surface on the atomic scale but still close to the substance on a macroscopic scale.

Work function depends only on the nature of the metal and is independent of the method of acquiring energy by the electron. Work function is measured in electronvolt. Alkali metals like sodium and potassium have work functions lower than that of other metals but, nowadays, for photoelectric emission, suitable alloys are mostly used.

**Demonstrative Experiment**

An evacuated glass bulb G with a quartz windowing is used. Through the window, a monochromatic beam of light is incident a plate T that can emit electrons. Plate T is generally coated. With an alkali metal (like sodium or potassium). When plate C is kept at a positive potential concerning T, it attracts photoelectrons emitted from T. Hence a current is set up in the circuit monochromatic.

This force acts, as a centripetal force for the revolving particle. qvB-Sg or, which can be recorded by the galvanometer G’. T is called photocathode and C is called anode. Rheostat Rh, in series with battery B, can be used to increase or decrease the potential difference V. Using the commutator C’, C can also be kept at negative potential concerning T.

**Ampere-Volt Characteristics Stopping Potential**

**Stopping Potential Definition:**

The minimum negative potential of anode concerning photocathode, for which photoelectric current becomes zero, is called stopping potential

**Keeping the frequency of light constant**

Graphs are drawn showing the dependence of I on V; where I = photoelectric current and V = potential difference between anode and cathode. Here a monochromatic light is used so that the frequency (f), of the incident light remains constant. 1 and 2 in this graph, represent I-V characteristics for different intensities of incident light.

**Detailed study of this graph reveals the following facts:**

**Saturation current:**Characteristic curves become horizontal for higher values of V. This shows that saturation current has been achieved. So, all the electrons, emitted by the photocathode, have been collected by the anode.**Effect of intensity of incident light:**At constant V, with a decrease in intensity i.e., the brightness of the incident, monochromatic light, the photoelectric current also decreases. Photoelectric current is directly proportional to the intensity of the incident light.**Stopping potential:**A When a negative potential is applied to anode concerning the photocathode, photoelectric Current docs do not show hut decreases gradually with an Increase In negative potential on C.

This Indicates that photoelectrons possess some Initial kinetic energy due to which they can reach the anode, overcoming the repulsive force of negative potential, With an Increase In the negative potential of the anode, the photoelectric current becomes zero ultimately. The negative potential at this stage Is called the stopping potential, cut-off potential, or cut-off voltage, VQ

**The value of stopping potential depends on two factors: **

- Nature of the surface of the photocathode and
- Frequency of the incident lightAnalysing graphs 1 and 2, we see that stopping potential VQ does not depend on the intensity of incident light.

An increase in the intensity of incident light only increases the t£p value of the saturation current

**Keeping the intensity of light constant**

Graphs are drawn showing the dependence of I on V; where = photoelectric current and V = potential difference between anode and cathode. Lights of different frequencies but of the same intensity are used as incident light on the cathode. In this figure, graphs 1 and 2 represent the I-V characteristic curves for different frequencies

In this case, as the frequency of incident light on the photocathode increases the y-value of stopping potential also increases and vice versa. But SEflyÿtion current is independent of the frequency of light

**Relation between kinetic energy of photoelectrons and stopping potential:**

When, the anode potential becomes equal to the stopping potential V_{0}, photoelectrons with even the highest kinetic energy, cannot reach the anode. Hence, the maximum kinetic energy of photoelectron (E_{max}) = loss of energy of electron for overcoming negative potential V_{0}.

When an electron of charge e overcomes a negative potential VQ, loss of energy of electron = eV_{0}. Thus,

E_{max }= eV_{0 }…………………….. (1)

Also, if the maximum initial velocity of the electron is v_{max}, then

⇒ \(E_{\max }=\frac{1}{2} m v_{\max }^2\)

m = Mass of electron

⇒ \(\frac{1}{2} m v_{\max }^2=e V_0\)

⇒ \(v_{\max }=\sqrt{\frac{2 e V_0}{m}}\)

The maximum kinetic energy of an electron Is independent of the Intensity of light

## Dual Nature Of Matter And Radiation Photoelectric Effect Numerical Examples

**Example 1. The stopping potential for monochromatic light of a metal surface is 4V. What is the maximum kinetic energy of photoelectrons?**

**Solution:**

From the relation, £max = eV_{0} we can say that, when the stopping potential is 4 V, maximum kinetic energy, E_{max} = 4eV.

**Example 2. For a metal surface, the ratio of the stopping potentials for two different frequencies of incident light is 1: 4. What is the ratio of the maximum velocities in the two cases? max **

**Solution:**

Stopping potential oc maximum kinetic energy. Again maximum kinetic energy ∝ (maximum velocity)². Hence, stopping potential ∝ . (maximum velocity)². If v_{1} and v_{2} are the maximum velocities in the two given cases, respectively then

⇒ \(\frac{1}{4}=\left(\frac{v_1}{v_2}\right)^2\)

i.e., v_{1}= v_{2}

= 1:2

**Threshold Frequency or Cut-off Frequency**

**Threshold Frequency Definition:**

The minimum frequency of incident radiation which can eject photoelectrons from the surface of a sub¬stance, is called the threshold frequency for that substance.

**Frequency versus stopping potential graph:**

In photoelectricity, neither the stopping potential nor maximum energy of photoelectrons is a constant quantity. Magnitudes of both V_{0} and E_{max} depend on

- The frequency of the incident light and
- The nature of the surface of the substance used.

The relation between frequency and stopping potential for different metals is shown graphically

**Characteristics of the graph:**

For each substance, there is a certain frequency f_{0 }of incident light, for which the maximum energy of photoelectrons becomes zero.

In other words, there is no emission of photoelectrons. Hence, whatever may be the intensity of incident radiation, no electron can leave the metal, surface for the light incident with a frequency equal to or less than f_{0} i.e., photoelectric emission stops. This f_{0} is the threshold frequency.

The maximum wavelength corresponding to the minimum frequency f_{0} is called the threshold wavelength. It is given by,

⇒ \(\lambda_0=\frac{c}{f_0}\) c= speed of light

**Discussions:**

- From what we see, the stopping potential or maximum kinetic energy increases as the frequency of incident radiation increases. Hence in practice, almost in all cases, ultraviolet rays are used, as the frequency of ultraviolet rays is much more than that of visible violet ray
- Alkali metals (For example, sodium, potassium, cesium, etc.) emit photoelectrons even for comparatively low-frequency light.

**Characteristics of Photoelectric Effect**

- Photoelectric current is directly proportional to the intensity of the incident light.
- The maximum velocity or kinetic energy of the photoelectron is independent of the intensity of incident light. On the other hand, maximum velocity or kinetic energy increases with an increase in the frequency of the incident light.
- For a given material, there exists a certain minimum frequency (threshold frequency, f
_{0}) of incident light bel which no photoelectrohs are emitted. Photoelectric effect is usually prominent in the range of frequencies of yellow, to ultraviolet radiation. - Threshold frequency is different for different materials. Photoelectrons emitted from the surface of a substance have any velocity between zero and maximum velocity.
- The emission of a photoelectron is an instantaneous process, which means that photoelectrons are emitted as soon as light falls on the metal surface. There is practically no time gap between these two incidents.
- The emission of photoelectrons makes the rest of the surface very slightly positively charged (this principle is followed in making photovoltaic cells). of electrons in, the photoelectric effect does not depend on the temperature of the surface

**Photoelectric Cell**

**Photoelectric Cell Definition:**

Cells, designed to convert light energy to electri¬ cal energy, based on the principle of the photoelectric effect, are photoelectric cells.

**Photo-voltaic cell: **

In this cell, an emf is developed directly from the photoelectric effect. This cell is an electric cell as it works as a source of EMF without any aid of an auxiliary cell.

**Description:**

A copper plate is taken and on one surface, a layer of cuprous oxide (Cu_{2}O) is deposited by the method of oxidation. Over this layer of Cu_{2}O, using the evaporation technique, a very thin coating of gold or silver is applied. This coating is so thin that light, especially ultraviolet rays, can easily penetrate it and reach the layer of Cu_{2}O.

**Working principle:**

When light Tails on Cu_{2}O, photoelectrons are emitted. These electrons instantaneously spread over the gold or silver coating. As a result, this coating achieves a negative potential concerning the copper plate, i.e., an effective emf is developed between the copper plate and the gold or silver coating. This emf sets up a current in the load resistance (R_{L}) used in the external circuit. This current is directly proportional to the intensity of incident light.

**Use of photoelectric cells:**

**1. Automatic switch:** In fire alarms, railway signals, streetlights, etc., automatic switches are made using photoelectric cells. QFJ Sound recording and reproduction: Photoelectric cells are used in sound recording and its reproduction, on the soundtrack of cinema and television.

**2. Solar cell:** A photo-voltaic cell is used to construct a solar battery. This battery is indispensable in spaceships and artificial satellites-

**3. Television camera:** In this device, the photoelectric cell is j used to convert optical images into video signals for television broadcasts.

**4. Automatic counting device:** A photoelectric cell is used to make an automatic counting device. The number of viewers entering or emerging from a hall can be counted with this device.

**5. Automatic camera:** A photoelectric cell is used in automatic cameras where the controlling of picture quality depends on the intensity of light.

**Failure of Wave Theory of Light**

The photoelectric effect cannot be explained by the concept of the wave theory of light. The following observations are in contradiction with the wave nature of light:

**Maximum kinetic energy of photoelectrons:**According to wave theory, energy carried by light waves increases with an increase in the intensity of light. Hence, when the material surface is illuminated with highly intense light, the kinetic energy of emitted photoelectrons will be very high and it will not have any upper limit. But from the value of stopping potential V_{0}, we see that the kinetic energy of the photoelectron cannot be more than e V_{0}.**Threshold frequency:**A highly intense light wave, even of a frequency less than threshold frequency f_{0}, carries a large amount of energy. This energy is sufficient to cause photoelectric emission. But no photoelectron is emitted in such cases. On the other hand, even a very low-intensity light beam of frequency greater than f_{0}can start photoelectric emission.**The photoelectric effect is instantaneous:**The energy carried by light waves incident on the surface of a substance needs a little time to be centralized in the limited space of an electron. Hence, there should be a time gap between the incidence of light waves on the surface and the emission of photoelectrons from the surface. But the photoelectric effect is instantaneous; that means there is no time gap between incidence and emission

## Dual Nature Of Matter And Radiation Quantum Theory of Radiation

**Photon:**

It has already been stated that the photoelectric effect cannot be explained in terms of the wave theory of light. In 1905, Einstein used Planck’s quantum theory and introduced the concept of photon particles. Thus, he could explain the photoelectric effect. The particle concept of radiation is the basis of quantum theory.

**The basic point of the theory is that electromagnetic** radiation is not a wave by nature but consists of a stream of particles called photons.

**Photons Properties:**

**The main properties of photons are:**

- Photons are electrically neutral.
- Photons travel with the speed of light, which does change under any circumstances, (velocity of light, c = 3 × 10
^{8}m . s^{-8}) - The energy carried by a photon, E = hf; where f = frequency of radiation and h = Planck’s constant. The energy radiated increases with the increased number of photons in its stream and hence the intensity of radiation also increases
- According to Einstein’s theory of relativity, the vast mass of a particle is zero, if it trawls at the speed of light Hence, the vast mass of each photon is zero.
- According to the theory of relativity, if the rest mass of a particle is m
_{0}(1 and its momentum is p, the energy of the particle,

E = \(\sqrt{p^2 c^2+m_0^2 c^4}\), In case of a photon, m_{0} = 0 ,

Hence E = pc, or p = \(\frac{E}{C}\) = hf/c. Thus, despite the photon being a massless particle, it has a definite momentum.

**Planck’s constant**

It is a universal constant

Unit of h in SI = \(\frac{\text { unit of } E}{\text { unit of } f}=\frac{\mathrm{I}}{\mathrm{s}^{-1}}\) = j- s

Unit of h in CGS system = erg .s.

This unit is the same as the unit of angular momentum. Thus h is a measure of angular momentum

Value of h = 6.625 × 10^{-34}J . s = 6.625 × 10^{-27}J . s

**Relation between the wavelength of radiation and the photon energy**

Energy of a photon, E = hf = \(\)

1 eV= 1.6. × 10^{-19}J . s and 1 A° = 10^{-10}m

Hence, expressing E in the eV unit and λ in the A° unit

λ A° = λ × 10^{-10} m EeV = E × (1.6 ×10^{-19}) J

Hence, E × (1.6 ×10^{-19}) J = \(\frac{\left(6.625 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{\lambda \times 10^{-10}}\)

Or, E = \(\frac{\left(6.625 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{\left(1.6 \times 10^{-19}\right) \times \lambda \times 10^{-10}} \approx \frac{12422}{\lambda}\)

Usually, the number on the right-hand side is taken as 12400.

Hence

E = 12400/ λ(inÅ) eV

## Dual Nature Of Matter And Radiation Quantum Theory of Radiation Numerical Examples

**Find the energy of a photon of wavelength 4950 A In eV ( h = 6.62 × 10 ^{-12} erg .s ). What is the momentum of this photon?**

**Solution:**

A = 4950 A = 4950 ×10^{-8} cm

Hence energy of a photon

E = hf \(\frac{h c}{\lambda}=\frac{6.62 \times 10^{-27} \times 3 \times 10^{10}}{4950 \times 10^{-8}}\)

= 4.012 × 10^{-12} erg

= \(\frac{4.012 \times 10^{-12}}{1.6 \times 10^{-12}}\)

= 2.5 eV

Momentum of photon

p = \(\frac{E}{c}=\frac{4.012 \times 10^{-12}}{3 \times 10^{10}}\)

= 1.34 × 10^{-22} dyn. s

**Example 2. Find the photons emitted per second by a source of power 25W. Assume, the wavelength of emitted light = 6000A°. h= 6.62 × 10 ^{-34} J.s.**

**Solution:**

λ = 6000 A° = 6000m × 10^{-10} ,c = 3 × 10^{8 }m.s^{-1}

Number of photons per shroud emitted by a source of power P Is,

n = \(\frac{p}{h f}=\frac{p}{h c / \lambda}=\frac{P \lambda}{h c}\)

n= \(\frac{25 \times 6000 \times 10^{-10}}{6,62 \times 10^{-34} \times 3 \times 10^8}\)

= 7.55 × 10^{19}

**Example 3. The wavelength of ultraviolet light Is 3 × 10 ^{-5}cm, What will be the energy of a photon of this light, in eV? (C = 3 ×10^{-10 }m.s^{-1} )**

**Solution:**

Planck’s constant, h = 6.625 × 10^{-27 } erg .s

Wavelength, λ = 3 × 10^{-5 }cm

The energy of a photon

E = \(\frac{h c}{\lambda}=\frac{\left(6.625 \times 10^{-27}\right) \times\left(3 \times 10^{10}\right)}{3 \times 10^{-5}} \mathrm{erg}\)

= \(\frac{\left(6.625 \times 10^{-27}\right) \times\left(3 \times 10^{10}\right)}{\left(3 \times 10^{-5}\right) \times\left(1.6 \times 10^{12}\right)} \mathrm{eV}\)

= 4.14 eV

**Example 4. Work functions of three metals A, H, and C arc 1.92 eV, 2.0 eV, and 5.0 eV respectively. Which metal will emit photoelectrons when a light of wavelength 4100A is Incident on the metal surfaces?**

**Solution:**

Energy of incident photon

E = hf

= \(h \cdot \frac{c}{\lambda}=\frac{6.625 \times 10^{-27} \times 3 \times 10^{10}}{4100 \times 10^{-8}} \mathrm{erg}\)

= \(\frac{6.625 \times 10^{-27} \times 3 \times 10^{10}}{1.6 \times 10^{-12} \times 4100 \times 10^{-8}} \mathrm{eV}\) ≈ 3.03 eV

Hence, this photon will be able to emit photoelectrons from metals A and B but not from C

**Example 5. In a microwave oven, electromagnetic waves are generated having wavelengths of the order of 1cm. Find the energy of the microwave photon. (h = 6.33 × 10 ^{-34 }J.s).**

**Solution:**

= 1 cm = 10^{-2} m; c = 3 × 10^{8} m .s^{-1}

E = hf

= \(\frac{h c}{\lambda}=\frac{\left(6.63 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{10^{-2}}\)

= 1.989 × 10^{-23} J

= \(\frac{1.989 \times 10^{-23}}{1.6 \times 10^{-19}}\) eV

= 1.24 × 10^{-4} 10 eV

## Dual Nature Of Matter And Radiation Einstein’s Photoelectric Equation

Einstein made the following assumptions to explain the photoelectric effect.

**Einstein’s postulates**

IQ A beam of light is incident on a metal surface as streams of photon particles. The energy of each photon having frequency is, E = hf (h = Planck’s constant).

Incident photons collide with electrons of metal. The collision may produce either of the two effects: The o photon gets reflected with its full energy hf or the Q photon transfers its entire energy hf to the electron.

Einstein used the quantum theory of radiation to explain the photoelectric effect.

**The entire energy hf of die incident photon, when transferred to an electron of the metal, is spent in two ways:**

- A part of the energy is spent to release the electron from the metal whose minimum value is equal to the work function WQ of the metal. But, due to the interaction of positive and negative charges inside a metal, most of the electrons need more energy than WQ for release.
- Rest of the energy, changes to the kinetic energy of released electrons. These moving electrons are photoelectrons that can set up photoelectric current. If energy absorbed by the electron to leave from the metal surface is the least i.e., WQ, the emitted electron attains maximum kinetic energy

Hence, hf = W_{0}+ E_{max}

E_{max}= hf – W_{0} ………………………..(1)

If the mass of an electron is m and the maximum velocity of a photoelectron is equation (1) we get,

½mv²_{max }= hf -W_{0 }……………………..(2)

Also, if V_{0} is the stopping potential for the incident light of frequency/, then Emax = eV0 (e = charge of an electron) Hence, from equation (1),

eV0 = hf-W_{0}

Equations (1), (2), and (3) are practically the same. Each of these is called Einstein’s photoelectric equation. In most collisions of photons with electrons, there is no energy transfer and the photons are reflected with their full energy hf. Hence the probability of photoelectric emission from the metal surface is low and the strength of photoelectric current never becomes very high

**Explanation of Photoelectric Effect by Quantum Theory**

Einstein’s photoelectric equation is based on the quantum theory of radiation.

**This equation correctly explains the following observations in the photoelectric effect:**

**1. The maximum kinetic energy of photoelectrons:**

Work function W_{0} is a constant for a fixed material surface; also the frequency of monochromatic light, f is a constant. Hence, E_{max} = hf- W_{0}, is also a constant. Thus, for fixed wavelength or fixed frequency of incident light, whatever may be the intensity, emitted photoelectrons cannot attain kinetic energy more than E_{max}

**2. Threshold frequency:**

The work function, JV0 is also a constant for a fixed material surface. If the frequency of incident light is decreased then as evident from equation E_{max}. = hf- W_{0}, the value of £max will come down to zero for a certain value of f = f_{0} (say).

0= hf_{0}– W_{0 }Or, hf_{0} = W_{0}

Or, f_{0 }= \(=\frac{W_0}{h}\) …………………………………(1)

If the value of f happens to be below f_{0}, the energy of the photoelectron turns out to be negative and there is no photoelectron emission. Hence, f0 is the threshold frequency. Putting W_{0 }= hf_{0} in Einstien’s photoelectric equation, E_{max } = hf – W_{0 },we get,

E_{max } = hf – hf_{0} = h( f- f E_{max } = h (f -f_{0} ) ………………………………(2)

Also, if λ and λ_{0} are the wavelength of the incident light and”threshold wavelength for the. metal surface respectively, then

f = c/λ and f_{0}= c/λ_{0}

c = Speed of light

Putting these values in equation (2), we get

E_{max}= \(h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)\) …………………………..(3)

Equations (2)equation. and (3) are the other forms of the arc of Einstein’s photoelectric equation.

**3. Photoelectric emission is instantaneous:**

Energy transfer takes place between a photon of energy hf and an electron In the metal due to their elastic collision. Hence, there is no delay in photoelectron emission after the incidence of light.

**4. Dependence of photoelectric current on the intensity of incident light:**

An increase in the intensity of incident light of a constant frequency increases the number of photons incident on the surface of the material.

Hence, the number of collisions between the photons and electrons increases. So, more electrons are emitted which increases photoelectric current This agrees with the results obtained experimentally

**Graphical representations of Einstein’s equation**

**1. Frequency (f) versus stopping potential ( V _{0}) graph:**

From equation (3) we have,

eV_{0 }= hf- W_{0}

Or, V_{0 }= \(\frac{h}{e} f-\frac{W_0}{e}\)

The graph obtained by plotting V_{0} against f is a straight line of the type, y = mx + c Knowing the charge of an electron e, Planck’s constant h can be calculated from the slope of the graph.

Work function W_{0} can be obtained from the Intercept \(\) in the y-axis. Also, the intercept with the x-axis gives the threshold frequency f_{0}.

It is important to note that the gradient of the straight Line Is \(\) for all substances but Intercepts from y- and x-axes, and f_{0}, respectively are different for different substances.

**2. Frequency (f) versus maximum kinetic energy (E _{max}) graph:**

From equation (1)

E_{0max }= hf – W_{0}

The graph obtained by plotting E_{max }against f is a straight line. Comparing the above relation with y = mx + c, we note

The slope of the graph is h, x -the axis Intercept is f_{0} and the y-axis intercept is – W_{0}.

## Dual Nature Of Matter And Radiation Einstein’s Photoelectric Equation Numerical Examples

**Example 1. The work function for zinc Is 3.6 eV. If the threshold frequency for zinc is 9 × 10 ^{14} cps, determine the value of Planck’s constant. (1eV = 1.6× 10^{-12} erg).**

**Solution:**

Work function, W_{0} = 3.6eV = 3.6 × 1.6 × 10^{-12} erg

Threshold frequency, f_{0} = 9 × 10^{14} cps = 9 × 10^{14 }Hz

As, W_{0} = hf_{0}

So, h = \(\frac{W_0}{f_0}=\frac{3.6 \times 1.6 \times 10^{-12}}{9 \times 10^{14}}\)

= 6.4 x 10^{-27} erg.s

**Example 2. The maximum kinetic energy of the released photoelectrons emitted from metallic sodium, when a light Is an Incident on It, is 0,73 eV. If the work function of sodium Is 1.82 eV, find the energy of the Incident photon In eV. Find the wavelength of incident light. (h = 6.63 × 10 ^{-27}erg s, eV = 1.6 × 10^{-12} erg )**

**Solution:**

From Einstein’s photoelectric equation, Emax hf – W_{0}, we get the energy of the incident photon as,

E = hf = E_{max} + W_{0} = 0.73 + 1.82 = 2.55 eV

Hence wavelength of incident light, \(\)

∴ λ = \(\frac{h c}{E}=\frac{6.63 \times 10^{-27} \times 3 \times 10^{10}}{2.55 \times 1.6 \times 10^{-12}}\) cm

λ = \(\frac{6.63 \times 10^{-27} \times 3 \times 10^{10} \times 10^8}{2.55 \times 1.6 \times 10^{-12}}\)A°

λ = 4875 A°

**Example 3. Light of wavelength 6000A° is Incident on. a metal. To. release an electron from the metal surface, 1.77 eV of energy Is needed. Find the kinetic energy of the fastest photoelectron. What is the threshold frequency of the metal (h = 6.63 × 10 ^{-27}erg s, eV = 1.6 × 10^{-12} erg ).**

**Solution:**

The energy of a photon,

hf = \(h \frac{c}{\lambda}=\frac{6.62 \times 10^{-27} \times 3 \times 10^{10}}{6000 \times 10^{-16}} \mathrm{erg}\)

= \(\frac{6.62 \times 10^{-27} \times 3 \times 10^{10}}{6000 \times 10^{-8} \times 1.6 \times 10^{-12}} \mathrm{eV}\)

= 2.07 eV

As per Einstein’s photoelectric equation,

E_{max } = hf – W_{0}= 2. 07- 1. 77 = 0. 3eV

Threshold frequency,

⇒ \(\frac{W_0}{h}=\frac{1.77 \times 1.6 \times 10^{-12}}{6.62 \times 10^{-27}}\)

= 4.28 × 10^{14} Hz

**Example 4. The photoelectric threshold wavelength for a metal is 3800A°. Find the maximum kinetic energy of the emitted photoelectron, when ultraviolet radiation of length 2000A° is incident on the metal surface. Planck’s constant, h = 6.62 × 10 ^{–}^{34} J s**

**Solution:**

Maximum kinetic energy of photoelectron,

E_{max }= hf-W_{0 }= hf-hf_{0} = \(\frac{h c}{\lambda}-\frac{h c}{\lambda_0}\)

= hc \(h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)=h c \frac{\lambda_0-\lambda}{\lambda \lambda_0}\)

In this case, the wavelength of the incident light,

λ = 2000 A° = 2000 × 10^{-10} m = 2 × 10^{-7} m

Threshold wavelength

λ_{0} = \(3800 \times 10^{-10} \mathrm{~m}=3.8 \times 10^{-7} \mathrm{~m}\)

Hence, E_{max } = \(\left(6.62 \times 10^{-34}\right) \times\left(3 \times 10^8\right) \times \frac{(3.8-2) \times 10^{-7}}{3.8 \times 2 \times 10^{-14}} \mathrm{~J}\)

= \(\frac{6.62 \times 10^{-34} \times 3 \times 10^8 \times 1.8}{3.8 \times 2 \times 10^{-7} \times 1.6 \times 10^{-19}} \mathrm{eV}\)

= 2.94 eV

**Example 5. The threshold wavelength for photoelectric emission from a metal surface is 3800 A. Ultraviolet light of wavelength 2600A° is incident on the metal surface, **

**Find the work function of the metal and****Maximum kinetic energy of emitted photoelectron.****( h= 6.63 × 10**^{–}^{27}erg.s )

**Solution:**

Threshold wavelength

λ_{0} = 3800 A° = 3800 × 10^{–}^{8} cm

∴ Work function,

W_{0} – hf_{0 }= \(\frac{6.63 \times 10^{-27} \times 3 \times 10^{10}}{3800 \times 10^{-8}} \mathrm{erg}\)

= \(\frac{6.63 \times 10^{-27} \times 3 \times 10^{10}}{3800 \times 10^{-8} \times 1.6 \times 10^{-12}} \mathrm{eV}\)

= 3.27 eV

As Per Einstein’s photoelectric equation, the maximum kinetic energy of photoelectron, E_{max} = hf-W_{0}

hf = kinetic energy of the incident photon

= hf_{0 } \(h f_0 \times \frac{f}{f_0}=h f_0 \frac{c / \lambda}{c / \lambda_0}=h f_0 \frac{\lambda_0}{\lambda}\)

= \(3.27 \times \frac{3800}{2600}\)

= 4.78

E_{max} = 4.78 – 3.27 = 1.51 eV

**Example 6. When radiation of wavelength 4940 A° is incident on a metal surface photoelectricity is generated. For a potential difference of 0.6 V between the cathode and anode, photocurrent stops. For another incident radiation, the stopping potential changes to 1.1V. Find the work function of the metal and wavelength of the second radiation. (**** h= 6.63 × 10 ^{–}^{27} erg.s , e = 1.6 × 10^{–}^{19} C)**

**Solution:**

For the first radiation, stopping potential V_{0} = 0.6V

∴ Maximum kinetic energy of photoelectron,

E_{max }= eV_{0 }= 0. 6eV

Wavelength, A = 4940 A° = 4940 × 10^{–}^{8 }cm

∴ The energy of an incident photon

= hf = \(h \frac{c}{\lambda}=\frac{6.6 \times 10^{-27} \times 3 \times 10^{10}}{4940 \times 10^{-8}}\)

= \(\frac{6.6 \times 10^{-27} \times 3 \times 10^{10}}{4940 \times 10^{-8} \times 1.6 \times 10^{-12}}\)

= 2.5 eV

If the work function of the metal is WQ, from Einstein’s equation

E_{max}= hf – W_{0}

Or, W_{0 }= hf – E_{max}

= 2.5 -0.6

= 1.9 eV

For the second radiation, V’_{0 }= 1.1V

Hence, E’_{max }= 1.1 eV

∴ E’_{max} = hf- W_{0}

Or, hf’ = E’_{max }+ W_{0} = 1.1 + 1.9

= 3.0 eV

Hence, \(\frac{h f}{h f^{\prime}}=\frac{2.5}{3.0} \text { or, } \frac{f}{f^{\prime}}=\frac{5}{6}\)

Or, \(\frac{c / \lambda}{c / \lambda^{\prime}}=\frac{5}{6} \quad \text { or, } \frac{\lambda^{\prime}}{\lambda}=\frac{5}{6}\)

Or, λ’ = \(\lambda \times \frac{5}{6}=4940 \times \frac{5}{6}\)

= 4117 A°(approx)

**Example 7. A stream of photons of energy 10.6 eV and intensity 2.0 W.m ^{–}^{2} is incident on a platinum surface. The area of the surface is 1.0 × 10^{–}^{4 }m^{2 }and its work function is 5.6 eV. 0.53% of incident photons emit photoelectrons. Find the number of photoelectrons emitted per second and the maximum and minimum energies of the emitted photoelectrons in eV. ( 1eV = 1.6 × 10^{–}^{19 }J)**

**Solution:**

If the intensity of incident light is I, the energy incident on a surface area A is IA. Hence, number of photons incident per second n = \(\frac{I A}{h f}\)

If x% of photons help to emit photoelectrons, the number of photoelectrons emitted per second,

N = \(n \times \frac{x}{100}=\frac{I A x}{h f \cdot 100}\)

Given, I = 2.0 W m^{–}^{2} , A = 1.0 × 10^{–}^{4} m^{–}^{2}

hf = 10.6 eV = 10.6 × 1.6 × 10^{–}^{19 }J and x = .0.53

N = \(\frac{2.0 \times 1.0 \times 10^{-4} \times 0.53}{10.6 \times 1.6 \times 10^{-19} \times 100}\)

= 6.25 × 10^{11}

Minimum kinetic energy of emitted photoelectron = 0

Maximum kinetic energy, E_{max} = hf- W_{0} = 10.6 – 5.6 = 5 eV

**Example 8. At what temperature would the kinetic energy of a gas molecule be equal to the energy of a photon of wavelength 6000A°? Given, Boltzmann’s constant, k = 1.38 × 10 ^{–}^{23}J K^{–}^{1}, Plank’s constant , h = 6.625 × 10^{–}^{34 }J. s**

**Solution:**

Let the required temperature be TK. We know, the kinetic energy of a gas molecule

= \(\frac{3}{2} k T=\frac{3}{2} \times 1.38 \times 10^{-23} \times T\)

Again, the kinetic energy of the photon

= \(h f=\frac{h c}{\lambda}=\frac{6.625 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-10}}\)

Hence,

= \(\frac{3}{2} \times 1.38 \times 10^{-23} \times T\)

= \(\frac{6.625 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-10}}\)

T = \(\frac{2}{3} \times \frac{6.625 \times 10^{-34} \times 3 \times 10^8}{1.38 \times 10^{-23} \times 6000 \times 10^{-10}}\)

= 1.6 × 10^{4 }k

**Example 9. The ratio of the work functions of two metal surfaces is 1: 2. If the threshold wavelength of the photoelectric effect for the 1st metal is 6000 A°, what is the corresponding value for the 2nd metal surface?**

**Solution:**

If the work function of 1st and 2nd metals be W_{0} and W’_{0} , respectively then

⇒ \(\frac{W_0}{W_0^{\prime}}=\frac{h f_0}{h f_0^{\prime}}=\frac{h c / \lambda_0}{h c / \lambda_0^{\prime}}=\frac{\lambda_0^{\prime}}{\lambda_0}\)

Or, \(\lambda_0^{\prime}=\lambda_0 \times \frac{W_0}{W_0^{\prime}}\)

= \(6000 \times \frac{1}{2}\)

= 3000 A°

**Example 10. The work function of a metal surface is 2 eV. The maximum kinetic energy of photoelectrons emitted from the surface for Incidence of light of wavelength 4140 A is 1 eV. What is the threshold wavelength of radiation for that surface**

**Solution:**

From Einstein’s photoelectric equation,

E_{max }= hf-W_{0}

Or hf = E_{max } + W_{0}= 1+2 = 3 eV

Now, \(\frac{h f}{W_0}=\frac{h f}{h f_0}=\frac{f}{f_0}=\frac{c / \lambda}{c / \lambda_0}=\frac{\lambda_0}{\lambda}\)

∴ λ_{0} = \(\lambda \frac{h f}{W_0}\)

= 4140 ×\(\frac{3}{2}\)

= 6210A°

**Example 11. The work function of a metal is 4.0 eV. Find the maximum value of the wavelength of radiation that can emit photoelectrons from the metal**

**Solution:**

Let, threshold frequency = f_{0} , threshold wavelength λ_{0} and work function = W_{0}

W_{0} =\(h f_0=\frac{h c}{\lambda_0} \text { or, } \lambda_0=\frac{h c}{W_0}\)

Given W_{0} = 4.0 eV = 4 × 1.6 × 10^{–}^{12} erg

We know, h = 6.60 × 10^{–}^{27} ergs. And

And c = 3 × 10^{10} cm .s^{–}^{1}

λ_{0} = \(=\frac{6.60 \times 10^{-27} \times 3 \times 10^{10}}{1.6 \times 10^{-12} \times 4} \mathrm{~cm}\)

λ_{0} = 3.09375 × 10^{–}^{5} cm≈ 3094 A°

**Example 12. The maximum energies of photoelectrons emitted by a metal are E _{1} and E_{2} when the incident radiation has frequencies f_{1} and f_{2} respectively. Show that the Planks comment h and the work function W0 of the metal are \(\) **

**Solution:**

According to Einstein’s photoelectric equation we here,

E= hf_{1} – W_{0 }…………………………………(1)

E= hf_{2}-W_{0} ……………………………..(2)

E_{1}– E_{2} = h(f_{1}-f_{2})

∴ h = \(\frac{E_1-E_2}{f_1-f_2}\) ………………………………….(3)

For E_{1 }> E_{2}

Again multiplying equation (1) by f2 and equation (2) by f1 we obtain.

E_{1}f_{2}= hf_{1}f_{2} – f_{2}W_{0} …………………………………..(4)

E_{2}f_{1}= hf_{1}f_{2} – f_{1}W_{0} ………………………………..(5)

Subtracting equation (5) from equation (4), we get

E_{1}f_{2} – Ff = f_{1}W_{0}-f_{2}W_{0}= W_{2}(f_{1}-f_{2})

∴ W_{0 }= \(\frac{E_1 f_2-E_2 f_1}{f_1-f_2}\)

**Example 13. A photoelectric source is illuminated successively by monochromatic light of wavelength λ and λ/2 calculates the work function of the material of the source. If the maximum kinetic energy of the emitted photoelectric in the second case is 3 times that in the first case.**

**Solution:**

We know the kinetic energy of emitted photoelectrons,

k = hf-W_{0} = \(\frac{h c}{\lambda}\) – W_{0}

In the first case,

K_{1} = \(\frac{h c}{\lambda}\) – W_{0 }= \(\frac{h c}{\lambda}\) – W_{0 }

In the second case,

K_{2} = \(\) – W_{0 }= \(\) – W_{0 }

It is given that, K_{2} = 3K_{1}

Or, 2hc/λ = W_{0 }

= 3(hc/λ – W_{0 })

2W_{0 } = hc/λ

∴ W_{0 }= hc/2λ

## Dual Nature Of Matter And Radiation Nature Of Radiation: Wave-Particle Duality

Electromagnetic radiations, if assumed to be streams of photon particles, can explain phenomena like photoelectric emission, blackbody radiation, atomic spectra, etc. However the theory fails to explain other optical phenomena like Interference, diffraction, polarisation, etc.

On die other hand, the wave theory of radiation can interpret these phenomena successfully. Hence, depending on the type of experiment, radiation sometimes behaves like waves and sometimes like a stream of particles. Thus, theory and particle theory are not contradictory but complementary to each other. ‘Ibis Is railed wave-particle duality

## Dual Nature Of Matter And Radiation Matter Wave

It has already been stated that radiation shows both wave nature and particle nature. But the fact that matter can show wave nature was unimaginable till 1924 when French physical Louis de Broglie put forward the theory that a stream of material particles may behave as a wave.

**Most probably the following reasons led him to such a conclusion:**

- Nature prefers symmetry. Hence, two physical entities, matter and energy must co-exist In symmetry,
- If radiation can have both particle and wave nature would also possess particle and wave nature,
- We know that a beam of light, which Is a wave, can transfer energy and momentum at different points of a substance, ‘similarly, a stream of particles can also transfer energy and momentum at different points of a substance. Therefore, this stream of particles may be a matter wave.

**de Broglie’s hypothesis:**

Matter also consists of waves. For a radiation of frequency f, the energy of a photon

E =hf Or, E = hc/λ

∴ c = fλ

Or, λ = \(\frac{h}{E / c}=\frac{h}{p}\) ………………………………… (1)

Where p is the momentum of the photon.

As per de Broglie’s hypothesis, equation (1) is also applicable to an electron or any other particle. In this case, A gives the wavelength of the electron (or particle) of momentum p and is known as the** de Broglie wavelength.**

Thus, substituting the values of Planck’s constant h and momentum of the particle p in equation (1), we get the de Broglie wavelength of the wave associated with the moving particle.

Hence a stream of any particle behaves like a beam of light, i.e., like a wave. The wave is known as the matter wave. The wavelength of this matter wave,

λ = \(\frac{h}{p}=\frac{h}{m \nu}\) ……………………………………. (2)

Where, m = mass of the particle, v = velocity of the particle, p = mv = momentum of the particle.

We can make the following inferences from the above relation connecting wavelength (a characteristic of the wave) And

**Momentum (a characteristic of the particle) : **

- If v = 0, then λ = ∞, it means the waves are associated with moving material particles only.
- The de Broglie wavelength does not depend on whether the moving particle is charged or uncharged. It means that matter waves are not electromagnetic waves because electromagnetic waves are produced from accelerated charged particles.
- If the mass m and the velocity v of the particle are large, the associated de Broglie wavelength becomes very small. If the momentum of the particle increases, the wavelength decreases.
- The wave nature and particle nature of any physical entity (matter or radiation) are mutually exclusive, i.e., if we consider the particle nature of radiation at any instant, the wave nature of radiation is to be excluded at that instant.

In 1927 **C J Davisson and H. Germer of Bell Telephone, **laboratories and George P Thomson of the University of Aberdeen, Scot¬ land, were able to show diffraction of electron streams and hence established experimentally the existence of matter waves

Hence, any moving stream of particle or matter exhibits interference, diffraction, and polarisation phenomena which can only be explained with wave theory.

The interference pattern, obtained by using a double slit type of experiment, using about 70000 moving electrons is shown in

## Dual Nature Of Matter And Radiation de Broglie Wavelength Of Moving Electron

Let an electron of mass m move with velocity v. Its de Broglie wavelength is given by

λ = h/mv

If the kinetic energy of the electron is K, then

K = ½ mv² Or, v = \(\sqrt{\frac{2 K}{m}}\)

So, de Broglie wavelength is

λ = \(\frac{h}{m \sqrt{\frac{2 K}{m}}}=\frac{h}{\sqrt{2 m K}}\) ………………………………. (1)

Equation (1) is the expression for the de Broglie wavelength ofa moving particle in terms of its kinetic energy.

Now suppose an electron is at rest. It is accelerated through a potential difference V. The kinetic energy acquired by the electron is K = eV; e = charge of the electron.

On substituting the value of K in equation (1), the de Broglie wavelength associated with the electron is given by

λ = \(\frac{h}{\sqrt{2 m e V}}\) ……………………… (2)

Putting the values of h,e,m in ‘equation (2) we have,

λ = \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times V}}\)

= \(\frac{12.27 \times 10^{-10}}{\sqrt{V}} \mathrm{~m}\)

= \(\frac{12.27}{\sqrt{V}}\)

**Wavelength of matter-wave**

Let the velocity of an electron (mass = 9.1 × 10^{10} cm .s^{–}^{1}), v = 10^{7} m. s^{–}^{1}. de Broglie wavelength of the electron

λ = \(\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{\left(9.1 \times 10^{-31}\right) \times 10^7}\)

= \(\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{\left(9.1 \times 10^{-31}\right) \times 10^7}\)

= 7.3 × 10^{-11} cm .s^{–}^{1}

= 0.73 A°

This wavelength is equivalent to the wavelength of X-rays.

1. Let the mass of a moving marble, m = 10 g = 0.01 kg, and its velocity, v = 10 m s-1. Then de Broglie wavelength of the marble

⇒ \(\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{0.01 \times 10}\)

= 6.63 × 10^{-33} cm .s^{–}^{1}m

The value of this wavelength is too small to be measured or to be observed by any known experiment Existence of such small wavelengths in electromagnetic radiation or any other real waves is still unknown to us

From the above discussions, we can infer that de Broglie’s hypothesis is of no use in the case of the macroscopic objects that we encounter in our daily lives.

The concept of matter waves is only Important In the case of particles of atomic dimensions.

## Dual Nature Of Matter And Radiation de Broglie Wavelength Of Moving Electron Numerical Examples

**Example 1. What is the de-Broglie- wavelength-related to an electron of energy 100 eV? (Given, the mass of the electron, m = 9.1 x 10 ^{-31} kg, e = 6.63 × 10^{-34} **

**J. s)**

**Solution:**

Let the velocity of the electron be v. Its kinetic energy,

½ mv² = E

Or, m²v² = 2mE

Or, mv = \(\frac{h}{m v}=\frac{h}{\sqrt{2 m E}}\)

de Broglie wavelength related to the electron,

λ = \(\frac{h}{m \nu}=\frac{h}{\sqrt{2 m E}}\)

= \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 100 \times 1.6 \times 10^{-19}}}\)

= 1.23 × 10^{-10} m = 1.23A°

**Example 2. Calculate the momentum of a photon of frequency 5 × 10 ^{13} Hz. Given h = 6.6 × 10^{-34} J.s and c= 3 × 10^{8} m.s-1**

**Solution:**

de Broglie wavelength \(\frac{h}{m v}=\frac{h}{p}\)

Momentum p = h/λ

Therefore, the momentum of the photon.

p = \(\frac{h}{\lambda}=\frac{h f}{c}=\frac{\left(6.6 \times 10^{-34}\right) \times\left(5 \times 10^{13}\right)}{3 \times 10^3}\)

= 1.1 × 10^{-28} kg.m.s^{-1}

**Example 3. The wavelength A of a photon and the de Broglie wavelength of an electron have the same value. Show that the energy of the photon \(\) kinetic energy of the electron. Here m, their usual meaning
Solution:**

The energy of the photon. F. = hf = hc/λ

The kinetic energy of the electron

E’ = \(\frac{1}{2} m v^2=\frac{p^2}{2 m}\)

Where , p= mv= momentum

As p = \(\frac{h}{\lambda}\) So, E’ = \(\frac{h^2}{2 m \lambda^2}\)

E’ = \(\frac{h^2}{2 m \lambda^2}\)

Or, E = \(\frac{E}{E^{\prime}}=\frac{h c}{\lambda} \cdot \frac{2 m \lambda^2}{h^2}=\frac{2 \lambda m c}{h}\)

E = \(\frac{2 \lambda m c}{h}\) E’

**Example 4. An electron and a photon have the same de Hrogilr wave¬ length λ = I0 ^{-10}m. Compare the kinetic energy of the electron with the total energy of the photon**

**Solution:**

The kinetic energy of an electron having mass m and velocity v, K = ½ mv²

Wavelength λ = h/mv

Or, v = h/m λ

∴ K =½ m. \(\frac{h^2}{m^2 \lambda^2}=\frac{h^2}{2 m \lambda^2}\)

The energy of the photon of wavelength λ, E \(\frac{h c}{\lambda}\)

∴ \(\frac{K}{E}=\frac{h^2}{2 m \lambda^2} \cdot \frac{\lambda}{h c}=\frac{h}{2 m \lambda c}\)

= \(\frac{6.6 \times 10^{-34}}{2 \times 9.1 \times 10^{-31} \times 10^{-10} \times 3 \times 10^8}\)

= 0.012<1

Hence, the kinetic energy- of the electron is lower than the total energy of the photon

**Example 5. Calculate the de Broglie wavelength of an electron of kinetic energy 500 eV.**

**Solution: **

The kinetic energy of an electron 500 eV means the electron is accelerated by the potential 500 V So, the de Broglie wavelength associated with the electron

λ = \(\frac{12.27}{\sqrt{500}}\)

= 0.55 A°

## Dual Nature Of Matter And Radiation Experimental Study Of Matter-Wave

**Davisson and Germer Experiment:**

We physicists C J Davisson and L H Germer first established the reality of matter waves. The experiment conducted by them in the year 1927 is described below.

**1. Davisson and Germer experiment:**

The electrons evaporated from the heated filament (F) after coming out of the electron gun are passed through a potential difference V. Thus, they acquire a high velocity and hence increased kinetic energy.

These electrons are then directed through a narrow hole and are thus collimated into a unidirectional beam. The kinetic energy of each single electron of this beam is eV.

Arrangements are such that this beam impinges normally on a specially hewn nickel single crystal. Because of this impact, the electrons may be scattered in different directions through different angles ranging from 10° to 90°. The incident and scattered beams are generally referred to as the incident ray and scattered ray.

A collector D capable of rotating around the crystal measures the intensity of the rays scattered in different directions. What is measured in the process is the number of electrons collected per second

**2. Davisson and Germer Observation:**

The intensity I obtained for the different values of the angle of scattering θ of the scattered ray is expressed using a polar graph where the scattered intensity is plotted as a function of the scattering angle.

**The convention in this respect is this: **

- The point of incidence on the crystal is taken as the origin O of the graph
- The direction OP opposite the direction of the incident ray PO is taken as the standard axis of the graph.

Suppose, for a stream of particles with definite energy, which is incident on the p crystal, the scattering angles are θ_{1}, θ_{2,} and the corresponding intensities I_{1}, I_{2……}.Hence, the line joining the points is represented by the polar coordinates (I_{1}, θ_{1} ), (I_{2}, θ_{2} ), …, etc.

Will be the polar Q graph indicating the result of the experiment. The polar coordinates of the points A and B are (I_{1}, θ_{1} ) and (I_{2}, θ_{2} ); that is, I_{1} = OA, I_{2} = OB, and θ_{1} = ∠AOP, θ_{2}= ∠BOP. In this figure, the line OABQ joining the points A, B… is the polar graph.

In this experiment, if the kinetic energy of the incident electrons is considerably low [as for the 40 eV electrons, then it is observed that the intensity I corresponding to changes in the angle of scattering from 10° to 90° keeps decreasing continuously, which does not indicate any characteristic property of the electron beam.

After this, Davisson and Germer noticed that when the kinetic energy of the incident electrons is gradually increased, a distinct hump appears at 44 eV at a scattering angle of about 60°. The hump becomes most prominent at 54 eV and a scattering angle of 50°. At still higher potentials, the hump decreases until it disappears completely at 68.

If the nickel crystal is compared with a diffraction grating, it is observed that if an X-ray of wavelength 1.65A is incident normally on the nickel crystal instead of the electron beam, then also, a peak representing maximum intensity will be obtained at 50°

The angle of scattering. On the other hand, when the de Broglie formula for matter wave is applied, the de Broglie wavelength of 54 eV electrons is λ = \(\frac{12.27}{\sqrt{54}}\) = 1.67A°

This striking similarity of the experimental value to the theoretical value according to de Broglie’s relation demonstrates the behavior of the electron stream as a matter wave. The experiment is termed an electron diffraction experiment as well

- The atoms of pure solids are arranged in crystal lattices of definite shapes. If an infinitely large number, say, 1020 or. more, if such crystals are aligned in a three-dimensional array to form a large piece of matter, then the specimen is called a single crystal. For example, the cubical grains of common salt thus formed are each a single crystal.
- Davisson and Germer used it for their experiment comparatively. slow-moving electrons with energy varying between 30 eV and 60 eV. Later on, G.P. Thomson successfully conducted a diffraction experiment with even high-energy electron beams; electrons of energy as high as 10 keV to 50 keV. In the subsequent years, the wave nature of other elementary particles, like the proton and the neutron, has also been established experimentally

**The nature of matter-wave**

According to de Broglie’s hypothesis, every moving particle can be represented as a matter wave, which, obviously, therefore, has to obey certain conditions

- Corresponding to a moving particle, a matter wave must also be a moving wave or progressive wave.
- The wave velocity must be equal to the particle velocity.
- Because the particle has a definite position at any time, the matter wave must be such that it can indicate the position of the particle at that time.

A pure sine curve cannot represent a matter wave, which means that no proper idea can be formed about the true nature of the matter wave from such curves. Let us suppose that a progressive wave moving in the positive x-direction is given by

Ψ = a sin (ωt – kx + δ) where a = amplitude, ω = angular velocity, fc= propagation constant; and δ = phase difference

- This wave extends from x = ∞ to -∞ with no dissipation or damping anywhere along the path. J fence, It can never Indicate the Instantaneous position of the moving partly, OH If it is assumed however that the matter wave Js analogs, to the pure sine wave, then it can be shown.
- That the velocity ofthe de Broglie wave turns out to be greater than the speed of light in a vacuum, which is Impossible contradicting Einstein’s special theory of relativity,

Moreover, a sine wave does not exist In nature. No real wave can extend from -∞ to +∞ without any damping. This implies that what we observe in reality is a wave group. For a practical example, consider the shape of waves formed when a stone is dropped in a pond. A wave group as shown comprising just a couple of wave crests and wave troughs is formed in the water and proceeds in circles over the water’s surface.

**Wave group or wave packet:**

If there is a superposition of two or more sine waves of different frequencies, then the waveform changes. If any such sine waves of continuously varying frequencies are superposed, the resultant wave that is formed has a general form like the one This is what is called wave group or wave packet.

**Characteristics of wave packets are**

This too is a progressive wave moving in a definite direction, in this case, along the + x-axis.

It is a localized wave, which means that it is limited within a rather small interval. Hence, such a wave group can indicate the possible instantaneous location of a moving particle.

It can be shown analytically that the velocity of such wave groups, \(\). It is called group velocity. Rigorous calculation shows that vg = v; that is, the group velocity is equal to the particle velocity. Thus properly constituted wave group or wave packet alone can correctly represent a matter wave.

**Wave function**

It has to be noted with particular care that a matter wave is neither an elastic wave like a sound wave nor an electromagnetic wave like a light wave. This is because an elastic wave is associated with the vibration of particles in an elastic medium, whereas an electromagnetic wave involves the simultaneous vibrations of the electric field and the magnetic field vectors.

In analogy, u matter wave is associated with a quantity known as wave function – Ψ. The origin and propagation ofthe matter wave are perfectly consistent with the vibrations of this Ψ – function concerning position and time.

It can be noticed that while the moving particle exists at a particular point at a particular moment of its motion, the correspond¬ ing matter wave occupies an extent of space at that moment, rather than be limited to a point.

It can be inferred from quantum mechanics that this property is an inherent property of matter waves; a wave packet is never concentrated at a single point. The probability of the particle existing at a particular point at a particular time is given by Ψ², the modulus squared off,

**Photon wave**

We have seen, even before discussing wave-particle duality, that electromagnetic radiation has a dual nature too; radiation is sometimes represented by waves, at other times in terms of photon beams. It is possible also to represent moving photons by a corresponding waveform as is done with a moving particle.

A single photon will naturally be represented by a wave group. The only difference here is that this (photon) wave packet must be an electromagnetic wave packet with group velocity in vacuum or air equal to the velocity of light.

## Dual Nature Of Matter And Radiation Experimental Study Of Matter-Wave Numerical Examples

**Example 1. Find de Broglie wavelength of the neutron at 127°C. Given, K = 1.38 × 10 ^{-23} ; h = J. mol^{–}^{1}.K^{–}^{1}. Plank’s constant, h = 6.626 × 10^{-34}J.s. mass of neutron, m = 1.66 × 10^{-27} kg.**

Kinetic energy of neutron, E = \(\frac{3}{2}\) kT

We know E = ½ mv²

Or, 2Em = m²v²

Or, mv = \(\sqrt{2 E m}=\sqrt{2 m \times \frac{3}{2} k T}=\sqrt{3 m k T}\)

de Broglie wavelength

λ = \(\frac{h}{m v}=\frac{h}{\sqrt{3 m k T}}\)

= \(\frac{6.626 \times 10^{-34}}{\sqrt{3 \times 1.66 \times 10^{-27} \times 1.38 \times 10^{-23} \times 400}}\)

= 1.264 × 10^{-10}m

**Example 2. Under what potential difference should an electron be accelerated to obtain electron waves of A = 0.6 A f Given, the mass of the electron, m = 9.1x 10 ^{-31} kg; Planck’s constant, h = 6.62 x 10^{-34} J. s**

**Solution:**

We know λ = ½mv² = eV

∴ mv= \(\sqrt{2 m e V} \text { or, } \lambda=\frac{h}{\sqrt{2 m e V}}\)

Or, λ² = \(\frac{h^2}{2 m e V} \text { or, } V=\left(\frac{h}{\lambda}\right)^2 \cdot \frac{1}{2 m e}\)

V = \(\left(\frac{6.62 \times 10^{-34}}{0.6 \times 10^{-10}}\right)^2 \times \frac{1}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}}\)

= 418.04 V

**Example 3. An a -particle and a proton are accelerated from rest through the same potential difference V. Find the ratio of Broglie wavelengths associated with them.**

**Solution:**

The kinetic energy of a particle of mass m.

E = \(\frac{p^2}{2 m}\)

Where p = \(\sqrt{2 m E}\)

So, de Broglie wavelength, λ = \(\frac{h}{p}=\frac{h}{\sqrt{2 m E}}\)

Now, if a particle of charge q is accelerated by applying potential difference V, then

E = qV

λ = \(\frac{h}{\sqrt{2 m q V}}\)

∴ λ ∝ \(\frac{1}{\sqrt{m q}}\) when V is constant

Hence \(\frac{\lambda_1}{\lambda_2}=\sqrt{\frac{m_2 \cdot q_2}{m_1 \cdot q_1}}\)

For proton and a -particle \(\frac{m_a}{m_p}=4, \frac{q_a}{q_p}\) = 2

∴ \(\frac{\lambda_p}{\lambda_a}=\sqrt{\frac{m_a}{m_p} \cdot \frac{q_a}{q_p}}\)

= \(=\sqrt{4 \times 2}\)

= 2 \(\sqrt{2}\)

Hence, the required ratio 2\(\sqrt{2}\):1

**Example 4. For what kinetic energy of neutron will the associated de Broglie wavelength be 1.40 x 10 ^{-10} m? The mass of a neutron is 1.675 x 10^{-27} kg and h = 6.63 x 10^{-34} J**

**Solution:**

If m is the mass and K is the kinetic energy of the neutron, the de Broglie wavelength associated with it is given by,

λ = \(\frac{h}{\sqrt{2 m K}}\)

Or, K = \(\frac{h^2}{2 m \lambda^2}=\frac{\left(6.63 \times 10^{-34}\right)^2}{2 \times 1.675 \times 10^{-27} \times\left(1.40 \times 10^{-10}\right)^2}\)

= 6.69 × 10^{-21} J

**Example 5. Find the wavelength of an electron having kinetic energy 10eV.(h = 6.33 × 10 ^{-34}J m_{e} = 9 × 10^{-31} kg**

**Solution:**

The kinetic energy of the electron,

E = ½ mv² = 10 eV = 10 × (1.6 x 10^{-19})J

m²v² = 2mE, or, momentum p = mv = \(\sqrt{2 m E }\)

The de Broglie wavelength of the electron,

λ = \(\frac{h}{\sqrt{2 m E}}=\frac{6.63 \times 10^{-34}}{\left[2 \times\left(9 \times 10^{-31}\right) \times\left(10 \times 1.6 \times 10^{-19}\right)\right]^{1 / 2}}\)

= 3.9 × 10^{-10 }m

= 3.9 A°

**Example 6. An α – particle moves In a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m². What is the de Broglie wavelength associated with the particle? ,**

**Solution:**

The radius of a charged particle rotating in a circular path in a magnetic field

R = \(\frac{m v}{B q}\) or, mv = RBq

The de Broglie wavelength associated with the particle,

λ = \(\frac{h}{m v}=\frac{h}{R B q}\)

Here, R = 0.83 cm = 0.83 × 10^{-2 }m, B = 0.25 Wb. m^{–}²

q = 2e = 2 ×1.6 × 10^{-19}C

[Since α – particle]

λ = \(\frac{6.6 \times 10^{-34}}{0.83 \times 10^{-2} \times 0.25 \times 2 \times 1.6 \times 10^{-19}}\)

= 0.01 A°

## Dual Nature Of Matter And Radiation Very Short Questions And Answers

**Question 1. Below a minimum frequency of light, photoelectric emis¬ sion does not occur Is the statement true or false?**

**Answer:** True

**Question 2. Above the threshold wavelength for a metal surface, even a light of low intensity can emit photoelectrons. Is the statement true?**

**Answer:** No

**Question 3. What will be the effect on the velocity of emitted photo¬ electrons if the wavelength of incident light is gradually**

**Answer:** The velocity of electrons will increase

**Question 4. Why are alkali metals highly photo-sensitive?**

**Answer:** Work functions of alkali metals are very low

**Question 5. Express in eV the amount of kinetic energy gained by an electron when it is passed through a potential difference of**

**Answer:** 100 eV

**Question 6. The photoelectric threshold wavelength for a metal is 2100 A. If the wavelength of incident radiation is 1800 A, will there be any emission of photoelectrons? **

**Answer:** Yes

**Question 7. Which property of photoelectric particles was discovered from Hertz’s experiment? **

**Answer:** Property of negative charge

**Question 8. Which property of photoelectric particles was measured**

**Answer:** Specific charge

**Question 9. What is the relation between the stopping potential VQ and the maximum velocity v _{max} of photoelectrons?**

**Answer:**

⇒ \(\left[v_{\max }=\sqrt{\frac{2 e V_0}{m}}\right]\)

**Question 10. How does the kinetic energy of photoelectrons change due to an increase in the intensity of incident light?**

**Answer:** No change

**Question 11. Give an example of the production of photons by electrons.**

**Answer:** X-ray emission

**Question 12. Give an example of the production of electrons by photons.**

**Answer:** Photoelectric effect

**Question 13. If the intensity of incident radiation on a metal surface is doubled what happens to the kinetic energy of the electrons emitted?**

**Answer:** No change in KE

**Question 14. Write down the relation between threshold frequency and photoelectric work function for a metal.**

**Answer**: \(\left[f_0=\frac{W}{h}\right]\)

**Question 15. Which property of light is used to explain the characteristics of the photoelectric effect?**

**Answer:** Particle (photon) nature

**Question 16. What is the rest mass of a photon?**

**Answer:** Zero

**Question 17. The wavelength of electromagnetic radiation is X . What is the energy of a photon of this radiation?**

**Answer: **E = hc/λ

**Question 18. The light coming from a hydrogen-filled discharge tube falls on sodium metal. The work function of sodium is 1.82 eV and the kinetic energy of the fastest photoelectron is 0.73 eV. Fine, the energy of incident light. **

**Answer:** 2.55 eV

**Question 19. In the case of electromagnetic radiation of frequency f, what is the momentum of the associated photon?**

**Answer:** hf/c

**Question 20. If photons of energy 6 eV are incident on a metallic sur¬ face, the kinetic energy of the fastest electrons becomes 4eV. What is the value of the stopping potential?**

**Answer:** 4V

**Question 21. The threshold wavelength of a metal having work function W is X . What will be the threshold wavelength of a metal having work function 2W?**

**Answer:** λ/2

**Question 22. The work function of a metal surface for electron emission is W. What will be the threshold frequency of incident radiation for photoelectric emission?**

**Answer: **W/h

**Question 23. What is the effect on the velocity of the emitted photoelectrons if the wavelength of the incident light is decreased?**

**Answer:** Velocity will increase

**Question 24. Two metals A and B have work functions 4eV and 10 eV respectively. Which metal has a higher threshold wavelength **

**Answer:** Metal A

**Question 25. Ultraviolet light is incident on two photosensitive materials having work functions W _{1} and W_{2}(W_{1} > W_{2}). In which case will the kinetic energy of the emitted electrons be greater? Why**

**Answer:**For the material of work function W

_{2}

**Question 26. Is it possible to bring about the interference of electrons? **

**Answer:** Yes

**Question 27. What type of wave is suitable to represent the wave associated with a moving particle?**

**Answer:** Wave Packet

**Question 28. The wavelength of a stream of charged particles accelerated by a voltage Vis A. What will be the wavelength if the voltage is increased to 4 V? **

**Answer: **λ /2

**Question 29. The de Broglie wavelength of an electron and the wave¬ length of a photon are equal and its value is A = 10 ^{-10}m. Which one has higher kinetic energy?**

**Answer:**Photon

**Question 32. An electron and a proton have the same kinetic energy. Identify the particle whose de Broglie wavelength would be**

**Answer:** Electron

**Question 36. The de Broglie wavelength of a particle of kinetic energy K is A. What would be the wavelength of the particle if its kinetic energy were K/4?**

**Answer:** 2 λ

**Question 37. The de Broglie wavelength associated with an electron accelerated through a potential difference V is A. What will be its wavelength when the accelerating potential is increased to 4 V?**

**Answer:** λ /2

**Question 38. What will be the kinetic energy of emitted photoelectrons if light of threshold frequency falls on a metal?**

**Answer:** Zero

**Question 39. What conclusion Is drawn from the Davisson-Germer experiment? **

**Answer:**

The Davisson-Germer experiment proves the existence of matter waves. It can be concluded from the experiment that any stream of particles behaves as waves

**Question 40. Name the phenomenon which shows the quantum nature of electromagnetic radiation. **

**Answer: **

The phenomenon which shows the quantum nature of electromagnetic radiation is the photoelectric effect

## Dual Nature Of Matter And Radiation Fill In The Blanks

**Question 1. Maximum kinetic energy of photoelectrons depends on __________________ of light used but does not depend on the of light **

**Answer:** Frequency, intensity

**Question 2. In the case of photoelectric emission, the maximum kinetic energy of photoelectrons increases with an increase in the _______________ of incident light**

**Answer:** Frequency

**Question 3. Lenard concluded from his experiment that the particles emitted in the photoelectric effect are _________________**

**Answer:** Electrons

**Question 4. In the case of photoelectric emission, the maximum kinetic energy of emitted electrons depends linearly on the _________________ of incident radiation**

**Answer:** Frequency

**Question 5. Davison and Germer experimentally demonstrated the existence of _____________________ waves**

**Answer:** Matter

## Dual Nature Of Matter And Radiation Assertion Type

**Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements,**

- Statement 1 Is true, statement 2 Is true; statement 2 Is the correct explanation for statement 1.
- Statement 1 Is true, statement 2 Is true; statement 2 Is not a correct explanation for statement 1.
- Statement 1 Is true, statement 2 Is false.
- Statement 1 Is false, and statement 3 is true.

**Question 1.**

**Statement 1:** Any light wave having a frequency less than 4.8 × 10^{14 }Hz cannot emit photoelectrons from a metal surface having work function 2.0 eV.

**Statement 2: **If the work function of a metal is W_{0} (in eV), then the maximum wavelength (in A°) of the light capable of initiating a photoelectric effect in the metal is \(\lambda_{\max }=\frac{12400}{W_0} \)

**Answer:** 1. Statement 1 Is true, statement 2 Is true; statement 2 Is the correct explanation for statement 1.

**Question 2.**

**Statement 1:** The energy of the associated photon max becomes half when the wavelength of the electromagnetic wave is doubled.

**Statement 2:** Momentum of a photon

= \(\frac{\text { energy of the photon }}{\text { velocity of light }}\)

**Answer:** 2. Statement 1 is true, and statement 2 Is true; statement 2 Is not a correct explanation for statement 1.

**Question 3.**

**Statement 1:** In the photoelectric effect the value of stopping potential is not at all dependent on the wavelength of the light

**Statement 2:** The maximum kinetic energy of the emitted photoelectron and stopping potential are proportional to each other.

**Answer:** 4. Statement 1 is false, and statement 3 is true.

**Question 4.**

**Statement 1:** The maximum surface velocity does not of increase the photo electron even if the intensity of the incident electromagnetic wave is increased.

**Statement 2:** Einstein’s photoelectric equation

½ mv²_{max }= hf – W_{0}

Where the symbols have their usual meaning

**Answer:** 1. 1. Statement 1 is true, statement 2 Is true; statement 2 Is the correct explanation for statement 1.

**Question 5.**

**Statement 1:** The stopping potential becomes double when the frequency of the incident radiation is doubled.

**Statement 2:** The work function of the metal and the threshold frequency of the photoelectric effect are proportional to each other.

**Answer:** 4. Statement 1 is false, and statement 3 is true.

**Question 6.**

**Statement 1:** If the kinetic energy of particles with different masses is the same, then the de Broglie wavelength ofthe particles is inversely proportional to their mass.

**Statement 2:** The momentum of moving particles is inversely proportional to their de Broglie wavelengths.

**Answer:** 4. Statement 1 is false, and statement 3 is true.

**Question 7.**

**Statement 1:** If a stationary electron is accelerated with a potential difference of IV, its de Broglie wavelength becomes 12.27 Å approximately.

**Statement 2:** The relation between the de Broglie length A and the accelerating potential V of an electron is given by λ = \(\frac{12.27}{V}\) Å

**Answer:** 3. Statement 1 Is true, and statement 2 Is false.

**Question 8.**

**Statement 1:** A moving particle is represented by a progressive wave group.

**Statement 2:** A pure sinusoidal wave cannot represent the instantaneous velocity or position of a moving particle.

**Answer:** 1. Statement 1 Is true, statement 2 Is true; statement 2 Is the correct explanation for statement 1.

**Question 9.**

**Statement 1:** The wavelength of 100 eV photon is 124 Å.

**Statement 2:** The energy of a photon of wavelength λ in Å is

E = \(\frac{12400}{\lambda}\) eV

**Answer:** 1. Statement 1 Is true, statement 2 Is true; statement 2 Is the correct explanation for statement 1.

**Question 10.**

**Statement 1:** A proton, a neutron, and an a -particle are accelerated by the same potential difference. Their velocities will be in the ratio of 1: 1: √2.

**Statement 2:** Kinetic energy, E = qV =½mv²

**Answer:** 4. Statement 1 Is false, and statement 3 is true.

## Dual Nature Of Matter And Radiation Match The Columns

**Question 1. Light of fixed intensity is incident on a metal surface.’ Match the columns in case of the resulting photoelectric effect**

**Answer:** 1-B, 2 – C, 3-D, 4-A

**Question 2.**

**Answer:** 1-A, 2-C, 3-D, 4-B

**Question 3. Several stationary charged particles are accelerated with appropriate potential differences so that their de Broglie wavelengths are the same**

**Answer:** 1-B, 2-D, 3-A, 4-C

**Question 4. The work function and threshold frequency of photoelecj trie emission of a metal surface are WQ and f respectively. Light of frequency/ is incident on the surface. The mass and charge of the electron are m and e respectively**

**Answer:** 1-B, 2-D, 3-C, 4-A