WBCHSE Class 12 Physics Notes For Artificial Transmutation Of Elements

Atomic Nucleus Artificial Transmutation Of Elements

We have already seen that radioactive elements are trans¬ formed into new elements. We also know that the identity of an element depends on its proton number. Therefore, if the proton number of an element can somehow be changed, the element is said to have undergone artificial transmutation.

Generally, artificial transmutation is brought about in two ways:

1. Nuclear reaction: Here the nucleus is bombarded with high-energy particles. Thereby the nucleus changes and forms the nucleus of a new element
2.  Artificial radioactivity: Often the transmuted nucleus formed by the process of nuclear reaction, is not a stable one and exhibits radioactivity and thereby decays to form a stable nucleus of another element, i.e., a transmuted that only the nuclear reaction is artificial and the subsequent disintegration of the unstable radioactive product is a natural phenomenon

Nuclear Reactions Definition:

Nuclear reaction is the process of transmuting elements by bringing a change in the nucleus, artificially

Energy condition of nude or reactions:

The binding energy per nucleon of a stable nucleus is about 8 MeV. Almost equal or more than this amount of energy is to be supplied from outside to bring about any change in the nucleus. Generally, high-energy particles are used to hit the nucleus to supply energy and to break up the nucleus.

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The sources of these high-energy particles are:

1. α, β, γ, and y-rays from radioactivity
2. Stream of high-energy particles resulting from nuclear reactions can be used to bring about further nuclear reactions.
3. Particles as projectiles from particle accelerators like cyclotron, betatrons, etc

Equations Of nuclear reactions:

In any nuclear reaction, a still or stationary substance called a target is hit by a stream of high-energy particles, each called a projectile.

When a project tile hits the target then either of the two things happens: 

1. Target and projectile both remain unaltered. This process is called scattering.
2. The nucleus of the target changes into the nucleus of another element. The impact produces one or more particles emerging with high energy. These particles are called emergent particles and the newly formed nucleus is called product nucleus. This entire process is called a nuclear reaction

Hence, denoting the nucleus ofthe target as X, the projectile as a, the product nucleus as Y, and the emergent particle as b, the nuclear reaction can be shown as

a + X →  Y + b ……………………………………… (1)

1. Conservation of mass number:

If A₁, A₂, A₃, and A₄ are mass numbers of a, X, Y, and b, then from equation (1)

A₁+ A₂ =  A₃ +A₄  ……………………………………… (2)

2. Conservation of atomic number: If  Z₁, Z₂, Z₃, and Z₄  are atomic numbers of a, X, Y, and b, then from equation

Z₁+ Z₂ = Z₃ +Z₄  ……………………………………… (3)

3. Mass-energy conservation:

Mass lost during the nuclear reaction, changes to energy as per Einstein’s massenergy equivalence. The released energy in a nuclear reaction is called Q -value of the reaction. Q -value for particles from the equation (1),

Q = [(M_a + M_x)-(M_y+M_b)]c²………………………….. (4)

The reaction is exoergic when the Q -value is positive and endoergic when the Q -value is negative. In the second case, the reaction cannot take place unless a threshold energy is provided to the projectile.

WBBSE Class 12 Artificial Transmutation Notes

Rutherford’s experiment: Proton as a nuclear particle:

Rutherford was the first to bring about the artificial transmutation of elements. His experimental arrangement is represented schematically

C is a container with a window W covered by a thin sheet of aluminum. F is a detector or film to record any emergent particle through W. R is a source of polonium that emits a -rays in the decay process. It is placed opposite to the window. Chamber C is filled with pure nitrogen gas.

The observations after a considerable period are:

• Chamber C shows the presence of oxygen gas, on chemical analysis.
• Photographic plate F is exposed and relevant detectors establish that the emerging rays through the window are streams of high-energy protons.

From this experiment, it is understood that the proton is one of the constituents of the atomic nucleus.

Detailed investigations showed that the proton and hydrogen nucleus were identical. Hence, a symbol for proton in a nuclear reaction is ₂H¹.

Equation of reaction: The nuclear reaction in the chamber C, can be represented by

₂He⁴ (α) + ₇N¹⁴ (nitrogen)  → ₈O¹⁷ + ₁H¹ (oxygen)

In short form, the reaction is often represented as N14(a, p)017 and the explanation is when N¹⁴  as (α, p) O¹⁷ bombarded with an α -particle (2He4), O17 (oxygen) is produced j and a proton is emitted.

A few ar-induced transformations:

₂He⁴  + ₁₃A1²⁷ →  ₁₄Si³⁰ + ₁H¹ or, Al²⁷(a, p) Si³⁰

₂He⁴ + ₅B¹⁰→ ₆C¹³ + ₁H¹ or, B¹⁰  (α, p) C¹³

₂He⁴  + ₁₉K³⁹ → ₂₀Ca⁴² + ₁H¹ or, K²³⁹(α, p) Ca⁴²

Discovery of neutron: Chadwick’s experiment

1. Bothe-Baker’s experiment:

These German scientists in 1930, observed that when Be (beryllium) is exposed to a stream of a -particles, highly penetrating uncharged rays, are emitted. Initially, these rays were considered y

2.  Curie-Joliot’s experiment:

In 1932, Irene Curie and her husband Frederic Joliot observed that when a blt&k of paraffin wax was placed on the path of the above-mentioned rays, high-energy protons were emitted. Emissions in Bothe and Becker’s experiment, if taken as 9-rays, cannot account for the source of the high-energy protons produced in this experiment.

3.  Chadwick’s Analysis :

James Chadwick, in the same year, repeated the experiment and put forward the explanation. His experimental arrangement.

 Chadwick assumed that:

• The rays emitted due to the impact of α -particles on Be nucleus, were not electromagnetic waves like y-rays but a stream of neutral particles. He named these particles neutron
• Paraffin wax contains hydrogen atoms and every nucleus of the atom is a proton. Due to the elastic collision between. the stream of neutrons and the protons, the proton stream resulted
• The ionization chamber helps In finding the energy and momentum of the proton released. Now, by applying the theory of elastic I collision the mass of the neutron can be obtained.

This experiment established a neutron as a fundamental particle that constitutes a nucleus.

Short Notes on Nuclear Reactions

Three things are inferred here:

• That neutron is electrically neutral. Its mass Is slightly greater than the mass of proton. So Its effective atomic number and mass number are 0 and l respectively. St) lit a nuclear reaction it is represented as ₀n¹
• A free neutron is not a stable particle. It undergoes a natural $ -decay and changes to a proton.

₀n¹ → ₀H¹ + _-1β¹ 

The half-life period for radioactivity is about 12 min.

The nuclear reaction in beryllium can be represented as

₂He⁴ + ₄Be⁹ →  ₆C¹² +  ₀n¹

Artificiahir induced Radioactivity

Definition:

When an unstable isotope of a stable element Is artificially formed and if the isotope exhibits natural radioactivity it is called artificial or induced radioactivity

Examples:

Carbon, sodium, and phosphorus are stable elements, and then: isotopes C¹², Na²³, and P³¹ are stable isotopes. When C¹⁴, Na²⁴, P³⁰ isotopes are produced artificially, they are found to be radioactive. They are generally called radioactive isotopes or radioisotopes. LikeC¹⁴ is called radiocarbon, Na²⁴ is known as radiosodium, and so on. The radioactive decay mode of these isotopes can be represented as’

₆C¹⁴ → ₇N¹⁴ + _-1β⁰ :  Half-life = 5600 y

_11Na²⁴ → ₁₂Mg²⁴ + _-1β⁰ :  Half-life = 15 h

₁₅P³⁰ → ₁₄Si³⁰ + _-1β⁰ :  Half-life = 2. 5 min

Positive  β decay or β⁺ decay:

As evident from the above radioactive decay of P³⁰, the emitted particle is ₊₁β⁰ or positron. Positrons have the same mass as electrons but are positively charged (+e). Except for the positive charge, the β -decay and positron emission are identical. So, it is called β⁺ decay. β⁺ decay is found only in artificial radioactivity.

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Characteristics of artificial radioactivity:

• Radioisotopes are produced from naturally stable elements.
• Decay mode of radioisotopes; Atomicis similar to natural radioactivity of radioactive substances. It follows the exponential law of – radioactivity.
• Natural radioactive substances generally have higher mass numbers. But radioisotopes may be lighter.
• Radioisotopes can exhibit α -decay, β -decay, γ -decay, or β⁺ decay, the last one being characteristic of artificial radioactivity only.
• The displacement rule for fi+ decay is that the mass number remains unaltered but the atomic number decreases by 1.

Discovery of Curie-Joliot:

Irine Curie and Frederic Joliot first discovered artificial radioactivity. Projecting a -particles from/polonium on an aluminum target, they identified positron particles mixed with emitted neutrons. The while emission of the neutron is easily explained using the nuclear reaction

₂He⁴ + ₁₃Al²⁷ → ₁₅P³⁰ + ₀n¹

The presence, of positrons remained unexplained.

They further observed that:

• On stopping projectile a from the source, the emission of neutrons is stopped but the emission of positrons continues.
• The rate of, a decrease of positron emission from Al -target excited by α -particles, is exponential, i.e., it obeys the decay law of natural radioactivity.
• They, therefore, concluded that isotope P³⁰ produced in the above nuclear reaction must be a radioactive isotope

Neutron Induced Nuclear Reactions

A neutron, when used as a projectile Utilising nuclear reactions offers distinct advantages compared to using -particles or protons as projectiles. As a neutral particle, the initial energy of a neutron is not very important because it does not need to overcome the electrostatic repulsion from the positive nucleus. While alpha particles or protons require considerable energy to break the Coulomb barrier and reach the nucleus, neutrons are optimal projectiles for nuclear reactions since they do not need to lose any energy to reach the nucleus.

Thermal neutron

Nuclear reactions brought about by itÿpns of energy of a few MeV are similar to reactions caused by using a or proton projectiles. But when a slow-retrieving neutron hits a target, a set of new types of nuclear reactions take place

Neutrons of kinetic energy of the order of 10^-2 eV are called 1 thermal neutrons. This order is the same as the kinetic energy of atomic or subatomic particles at room temperature. Since, the external manifestation of the kinetic energy of atomic or subatomic particles is the thermal energy, these are called thermal neutrons. Thermal neutrons are very slow-moving neutrons.

Common Questions on Artificial Transmutation

U-238 and U-235

Two isotopes of uranium are present in the ore of natural uranium. These are ₉₂U²³⁵ and  ₉₂U^238, or, U – 238 and U- 235 for short. Their abundance in nature is in the ratio 140: 1 (99.28%:’0.7%).

Transuranic elements:

The atomic number of uranium is 92 and no element of an atomic number higher than 92 is found in nature. But when U-238 is hit by a thermal neutron then,

• The neutron is absorbed by U-238 and
• U-239 is formed and pi] U-239 undergoes βdecay and forms an isotope of atomic number 93, which is a new element and does not exist naturally. This artificially made element is called neptunium
  (Np)

1. ₀n¹ +  ₉₂U²³⁸  → ₉₂U²³⁹

2. ₉₂U²³⁹  → ₉₃U²³⁹  + _-1β⁰

Proceeding almost in a similar way, it has been possible to produce elements of atomic numbers 94, 95, 96, ……, and 118 in the laboratory. Elements of atomic number higher than 92 and produced artificially are called transuranic elements. Out of these elements, plutonium (₉₃Pu²³⁹ ) has the maximum practical use.

List of a few natural and artificial isotopes, decay mode, and half-life period:

Note that associated ϒ -radiation

Atomic Nucleus Artificial Transmutation Of Elements Numerical Examples

Example 1. Complete the following- nuclear reaction: ₁₃Al²⁷ + ₂He⁴ → ₁₅P³⁰  +?
Solution:

Let A and Z be” the mass and atomic number of the unknown particle response respectively. From the law of conservation of mass number

27 + 4 = 30 + A _ or, A= 1

From the law of conservation of atomic number

13 + 2 = 15 + Z or, Z = 0

The article is therefore a neutron ₀n¹

∴ The complete equation ofthe reaction

₁₃Al²⁷ + ₂He⁴ →  ₁₅P³⁰  + ₀n¹

Example 2. Complete-the following-nuclear- -reaction: ₇N¹⁴ + ₂He⁴ →  ₈O¹⁷  + ?
Solution:

According to the law of conservation of mass number,

14 + 4 = 17 +1

Again, according to the law of conservation of atomic number

7 + 2 = 8 +1

∴ The mass number ofthe unknown particle = 1 and its atomic number = 1.

So it is a proton

∴ The complete equation of the reaction:  

₇N¹⁴ + ₂He⁴ →  ₈O¹⁷ + ₁H¹

Practice Problems on Artificial Transmutation

3. Example, Identify the missing particle in the following two reactions

1. ₉F¹⁹ + ₁H¹ →  ₈O¹⁶+ ?
2. ₁₂Mg²⁵ + ? →  ₁₁Na²² → + ₂He⁴ 

Solution:

Since, 19 + 1 = 16 + 4, and 9 + 1 = 8 + 2, the mass number of the missing element = 4 and atomic number = 2 . As Z = 2, the element is α -particle.

The complete equation of the reaction:

₉F¹⁹ + ₁H¹ →  ₈O¹⁶+ ₂He⁴ 

As, 25 + 1 = 22 + 4 and 12 +1 = 11 + 2 , the mass number of the missing element = 1 and atomic number = 1 .

So, the element is a proton.

∴ The complete equation of the reaction:

₁₂Mg²⁵ + ₁He¹  →  ₁₁Na²² → + ₂He⁴ 

Example 4. When ₄Be⁹  is hit by α-particles of a new neutron element. is Identify emitted the element and write the complete reaction equation
Solution:

α -particle: ₂He⁴ ; neutron: ₀n¹ 

∴ Let the new element, be _ZX^A

From the laws of conservation of mass number and atomic number,

9 + 4 = A + 1 of, A = 12 and 4 + 2 = Z + 0 or, Z = 6

∴ The new element = ₆C¹² (carbon):

∴  The complete equation of the reaction:

₄Be⁹  + ₂He⁴ → ₆C¹² + ₀n¹ 

Important Definitions in Artificial Transmutation

Example 5. When an aluminum nucleus (₁₃Al²⁷ ) is hit by a proton a new element is formed with the emission of α -particle

1. Write the complete equation of the reaction
2. Identify the new element and
3. Determine the number of neutrons and protons in the nucleus.

Solution:

Proton: ₁H¹; α  -particle: ₂He⁴

Let the new element =_ZX^A

[where A = mass number; Z = atomic piifribesr]

From the laws of conservation of mass and atomic number,

27 + 1 = A + 4 and 13 +1 = Z + 2 .

or, A = 24 and Z = 12

1. The complete equation of the reaction:

₁₃Al²⁷ + ₁H¹  →  ₁₂Mg²⁴  + ₂He⁴ 

2. As Z = 12, the element is magnesium (Mg)

So, the new element = ₁₂Mg²⁴

3. Now, we know, mass number = proton number ,…+ neutron number (x)

Or, 24 = 12 + x [  Since proton number = atomic number]

or x = 12

Example 6. On collision with a neutron, ₃Al²⁷  changes to radiOsodium ₁₁Na²⁴ and emits a particle. ₁₁Na²⁴, in its turn, emits a particle and is transmuted to ₁₂Mg²⁴ Write the two nuclear equations and identify the particles.
Solution:

Let the first equation of the reaction be

₁₃Al²⁷ + ₀n¹  →  ₁₁Na²⁴  +  _ZX^A  …………………………. (1)

The second equation of the reaction be

₁₁Na²⁴→  ₁₂Mg²⁴+ \(z_1 X^{A_1}\)  . …………………………. (2)

Applying the laws of conservation of atomic number and mass number from equation (1), 27 + 1 = 24+A, or, A = 4 and

Also, from equation (2), 24 = 24 + A₁ or, A₁ = 0  and 11 = 12 + Z₁ or,  Z₁ =  -1

Hence, the particles are helium nuclei i.e., α -particle in equation (1) and β -particle in equation (2).

The complete equations are

₁₃Al²⁷ + ₀n¹  →  ₁₁Na²⁴  + ₂He⁴ 

And ₁₁Na²⁴   →  ₁₂Mg²⁴ + _-1β⁰

Examples of Artificial Transmutation Reactions

Example 7. A nucleus disintegrates into two nuclei and their velocity and cities are in the ratio of 2: 1 . What will be the ratio of their sizes? 
Solution:

From the law of conservation of momentum, the two nuclei will have- the same magnitude of momentum. Hence, mass ∝ \(=\frac{1}{\text { velocity }}\) . Again, mass∝ R³, where R is the radius.

⇒ \(\text { mass } \propto R^3 \propto \frac{1}{\text { velocity }}\)

⇒  \(\frac{R_1}{R_2}=\left(\frac{v_2}{v_1}\right)^{\frac{1}{3}}=\left(\frac{1}{2}\right)^{\frac{1}{3}}\)

R₁ : R₂ = 1: 2^1/3

Example 8. In the nuclear reaction X(n, α)₃Li⁷ , identify X-
Solution:

The equation of the reaction:

_ZX^A   + ₀n¹  →  ₃Li⁷  + ₂He⁴ 

Using the conservation laws,

A + 1 = 7 + 4 or, A = 10

And Z + 0 = 3 + 2 or, Z = 5

These are the Z and A values of boron (B) .

The unknown element, X = ₅B¹⁰  (boron nuclide)