## Atom Multiple Choice Questions

**Question 1. The binding energy of an H atom, considering an electron moving around a fixed nucleus (proton), is ****B = \(\frac{m e^4}{8 n^2 \epsilon_0^2 h^2}\) ( m = mass of an electron) ****If one decides to work in a frame of reference where the electron is at rest, the proton would be moving around it. By similar arguments, the binding energy would be ****B = \(\frac{m e^4}{8 n^2 \epsilon_0^2 h^2}\) ( m = mass of proton). ****This last expression is not correct because**

- n would not be integral
- Bohr’s quantization applies only to electrons
- The frame in which the electron is at rest Is not inertial
- The motion of the proton would not be in circular orbits, even if approximately

**Answer:** 3. The frame in which the electron is at rest Is not inertial

**Question 2. The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because**

- Of the electrons not being subject to a central force
- Of the electrons colliding with each other
- Of screening effects
- The force between the nucleus and an electron will no longer be given by Coulomb’s law

**Answer: ** 1. Of the electrons not being subject to a central force

**Question 3. O _{2} molecules consist of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms**

- Is not important because nuclear forces are short-ranged
- Is as important as electrostatic force for binding the two atoms
- Cancels the repulsive electrostatic force between the nuclei
- Is not important because oxygen nuclei have an equal number of neutrons and protons

**Answer:** 1. Is not important because nuclear forces are short-ranged

**Read and Learn More Class 12 Physics Multiple Choice Questions**

**Question 4. For the ground state, the electron in the H atom has an angular momentum =ti, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. It is not true,**

- Because the Bohr model gives incorrect values of angular momentum
- Because only one of these would have a minimum energy
- Angular momentum must be in the direction of the spin of electrons
- Because electrons go around only in horizontal orbits

**Answer:** 1. Because the Bohr model gives incorrect values of angular momentum

**Question 5. Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is**

- 10.20 eV
- 20.40 eV
- 13.6 eV
- 27.2 eV

**Answer: ** 1. 10.20

**Question 6. An ionized H molecule consists of an electron and two protons. The protons are separated by a small distance of the order of angstrom. In the ground state**

- The electron would not move in circular orbits
- The energy would be 24 times that of a H -atom
- The electrons, orbits would go around the protons
- The molecule soon decays in a proton and a neutron

**Answer:** 1 and 3

**Question 7. Consider aiming a beam of free electrons towards free protons. When they scatter, an electron, and a proton cannot combine to produce a H atom**

- Because of energy conservation
- Without simultaneously releasing energy in the form of radiation
- Because of momentum conservation
- Because of angular momentum conservation

**Answer:** 1 and 2

**Question 8. The Bohr model for the spectra of a H atom**

- This will not apply to hydrogen in the molecular form
- Will not be applicable as it is for a He atom
- Is valid only at room temperature
- Predicts continuous as well as discrete spectral lines

**Answer:** 1 and 2

**Question 9. The simple Bohr model does not apply to the He atom because **

- He
^{4}is an inert gas - He
^{4}has neutrons in the nucleus - He
^{4}has one more electron - Electrons are not subject to central forces

**Answer:** 3 and 4

**Question 10. The Balmer series for the H atom can be observed**

- If we measure the frequencies of light emitted when an excited atom falls to the ground state
- If we measure the frequencies of light emitted due to transitions between excited states and the first excited state
- In any transition in an H atom
- As a sequence of frequencies with the higher frequencies getting closely packed

**Answer:** 2 And 4

**Question 11. Let \(E_n=-\frac{1}{8 \epsilon_0^2} \frac{m e^4}{n^2 h^2}\) be the energy of the nth level of the H atom If all the H atoms are in the ground state and radiation of frequency (E _{2 }– E_{1})/h falls on it**

- It will not be absorbed at all
- Some atoms will move to the first excited state
- All atoms will be excited to the n = 2 state
- No atoms will make a transition to the n = 3 state

**Answer:** 2 And 4

**Question 12. According to Rutherford’s atomic model, which of the following is correct?**

- Atom is stable
- The majority of space in an atom is empty
- E = hf
- None of these

**Answer:** 2. The majority of space in an atom is empty

**Question 13. Which of the following was used in Rutherford’s experiment**

- Aluminium
- Platinium foil
- Silver foil
- Gold foil

**Answer:** 4. Gold foil

**Question 14. What is the velocity of the electron in the first Bohr orbit of a hydrogen atom?**

- 3 × 10
^{8}m.s^{-1} - 2.19 × 10
^{6}m. s^{-1} - 3 × 10
^{7}m. s^{-1} - 3 × 10
^{7}m. s^{-1}

**Answer:** 2. 2.19 × 10^{6} m. s^{-1}

**Question 15. How many times will the electron in the first Bohr orbit of a hydrogen atom revolve in 1 s?**

- 6.58 × 10
^{15} - 4. 13 × 10
^{16} - 1.64 × 10
^{15} - 4.13× 10
^{15}

**Answer:** 1. 6.58 × 10^{15}

**Question 16. The total energy of an electron in the ground state of a hydrogen atom is**

- Zero
- 13.6 eV
- – 13.6 eV
- -13.6 J

**Answer:** 3. – 13.6 eV

**Question 17. What is the minimum energy (in eV) necessary to liberate an electron from the ground state of a Li ^{++} ion (Z = 3), according to Bohr’s theory?**

- 1.51
- 13.6
- 40.8
- 122.4

**Answer:** 4. 122.4

**Question 18. An electron in a hydrogen atom undergoes a transition from n _{1} to n_{2}, where n_{1} and n_{2} are the principal quantum numbers of two given states. According to Bohr’s theory, if the period ofthe electron in the initial state is eight times that in the final state, the possible values of n_{1 }and n_{2} will be, respectively**

- 4 and 2
- 8 and 2
- 8 and 1
- 6 and 4

**Answer:** 4 and 2

**Question 19. The atomic number of Helium is 2. What is the energy of the ground state of Helium ion having a single positive charge?**

- -13.6 eV
- -54.4 eV
- -40.8 eV
- -122.4 eV

**Answer:** 2. -54.4 eV

**Question 20. The ionization potential of an atom is 24.6 V. How much energy is required to ionize it?**

- 24.6 eV
- 2.46 eV
- 246 eV
- 0.246 eV

**Answer:** 1. 24.6 eV

**Question 21. In Bohr’s model of the hydrogen atom**

- The radius of the n-th orbit is directly proportional to n²
- The total energy of an electron in the n-th orbit is inversely proportional to n
- The angular momentum of an electron in any orbit is an integral multiple of h
- The potential energy of an electron in any orbit is more than its kinetic energy

**Answer:** 1. The radius of the n-th orbit is directly proportional to n²

**Question 22. The ground state energy of the hydrogen atom is -13.6 eV. If the electron in this atom jumps from the fourth level to the second level, what will be the wavelength of the emitted radiation?**

- 2918 Å
- 1824 Å
- 4863 Å
- 3824 Å

**Answer:** 3. 4863 Å

**Question 23. The electron in a hydrogen atom has been excited to the nth state. What is the maximum number of spectral lines that may be emitted by the atom when the electron transits to the ground state?**

- \(\frac{1}{6} n(n-1)(n-2)\)
- n
- n(n-1)
- \(\frac{1}{2} n(n-1)\)

**Answer:** 4. \(\frac{1}{2} n(n-1)\)

**Question 24. If the electron in a hydrogen atom is raised to the third orbit, how many photons of different energies may be emitted?**

- 1
- 2
- 3
- 4

**Answer:** 3. 3

**Question 25. According to Bohr’s model, the ratio of the energies of the electron in the first orbit of the hydrogen atom and the He ^{+} atom is**

- 1:2
- 4:1
- 1:4
- 1:9

**Answer:** 3. 1:4

**Question 26. The quantized physical quantity of an atomic electron, according to Bohr’s model, is**

- Linear velocity
- Angular velocity
- Linear momentum
- Angular momentum

**Answer:** 4. Angular momentum

**Question 27. A one-electron atom has an energy of -3,4eV. The kinetic energy of the electron is**

- -3.4 eV
- 3.4 eV
- -6.8 eV
- 6.8 eV %

**Answer:** 2. 3.4 eV

**Question 28. A one-electron atom has an energy of -3.4 eV. The potential energy of the electron is **

- -3.4 eV
- 3.4 eV
- -6.8 eV
- 6.8 eV

**Answer:** 3. -6.8 eV

**Question 29. An electron is in the third orbit of a hydrogen atom. If h = 6.6 × 10 ^{-34}J.s, its orbital angular momentum is**

- 1.98 × 10
^{33}J.s - 2.2 × 10
^{-34}J.s - 3.15 × 10
^{-34}J.s - 1.05 × 10
^{-34}J.s

**Answer:** 3. 3.15 × 10^{-34}J.s

**Question 30. The ratio between the radii of the fourth and the second electron orbits of a hydrogen atom is**

- 2:1
- 4:1
- 8:1
- 16:1

**Answer:** 2. 4:1

**Question 31. The ratio between the electron velocities in the second and the third orbits of a hydrogen atom is**

- 4:9
- 2:3
- 3:2
- 9:4

**Answer:** 3. 3:2

**Question 32. The ratio of the electron revolution frequencies in the first and in second orbits of a hydrogen atom is**

- 8:1
- 4:1
- 2: 1
- 1:4

**Answer:** 8:1

**Question 33. In an inelastic collision, an electron excites as a hydrogen atom from its ground state to a M -shell state. A second electron collides instantaneously with the excited hydrogen atom in the M -state and ionizes it. At least how much energy does the second electron transfer to the atom in the M – state?**

- + 3.4 eV
- +1.51 eV
- -3.4 eV P
- -1.51 eV

**Answer:** 2. +1.51 eV

**Hint:** Total energy of the electrons in n -th state hydrogen of the atom,

⇒ \(E_n=-\frac{13.6}{n^2} \mathrm{eV}\)

For M state

E_{m} = \(-\frac{13.6}{3^2}\)

= 0 – (-1.51)= -1.51 eV

∴ The least energy transferred to the atom

= -1.51 eV

**Question 34. The energy required for the electron excitation In Li ^{2+} from the first to the thin! Bohr orbit Is**

- 36.3 eV
- 108.8 eV
- 122.4 eV
- 12.1 eV

**Answer:** 2. 108.8 eV

**Hint:** Total energy of the nth state of an atom

⇒ \(-13.6 \frac{Z^2}{n^2} \mathrm{eV}\)

For Li^{2+} Z = 3

**Question 35. The wavelength of the first line of the Lyman series for a hydrogen atom is equal to that of the second line of the Balmer series for a hydrogen-like ion. The atomic number Z of hydrogen-like ion is**

- 3
- 4
- 1
- 2

**Answer:** 4. 2

**Hint: **The wavelength of the first line of the Lyman series of hydrogen atoms is given by

⇒ \(\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\)

Or, \(\lambda=\frac{4}{3 R}\)

The wavelength of the second line of the Balmer series for hydrogen-like atoms is given by

⇒ \(\lambda=\frac{4}{3 R}\)

According to the question

⇒ \(\frac{1}{\lambda^{\prime}}=Z^2 R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\)

⇒ \(\lambda^{\prime}=\frac{16}{3 Z^2 R}\)

Or Z =2

According to the question

Or = \(\frac{4}{3 R}=\frac{16}{3 Z^2 R}\)

Or, Z = 2

**Question 36. An electron in the hydrogen atom jumps from the excited state n to the ground state. The wavelength so emitted indicates a photosensitive material having a work function of 2.75 eV. If the stopping potential of the photoelectron is 10V, then the value of n is**

According to Einstein’s photoelectric equation

eV_{0} = hf- W_{0}

Or, hf – eV_{0} +W_{0}

= 10+ 2.75

= 12.75 eV

When an electron In the hydrogen atom makes a transition from excited state n to the ground state (n = 1), then the frequency (f) of the emitted photon is given by,

hf = \(E_n-E_1\)

=\(-\frac{13.6}{n^2}-\left(-\frac{13.6}{1^2}\right)\)

Or, n= 4

**Question 37. Out of the following which one is not a possible energy for a photon to be emitted by a hydrogen atom according to Bohr’s atomic model?**

- 0.65 eV
- 1.9 eV
- 11.1 eV
- 13.6 eV

**Answer:** 3. 11.1 eV

**Hint:** The energy of n th orbit of a hydrogen atom is given by,

⇒ \(E_n=-\frac{13.6}{n^2}\)

∴ E_{∞} – E_{1} = 0- (13.6 ) = 13.6 eV

E_{3} – E_{2} = = -1.5 -(-3.4) = 1.9 eV

E_{4 }– E_{3} = 0.85 – (1.5 ) eV = 0.65 eV

**Question 38. The wavelength λ and the frequency f of a particular X-ray spectral line varies with the atomic number Z of the target element. In this event, Z is nearly proportional to**

- λ
- √λ
- f
- √f

**Answer:** 4. √f

**Question 39. The ratio of the wavelengths of Ka and KQ spectral lines of hydrogen is**

- 8: 27
- 16: 27
- 27: 32
- 9: 16

**Answer:** 3. 27: 32

**Question 40. Let, λ _{21}, and λ_{31 }and λ_{32 }be the wavelengths of the spectral lines for the transition of electrons in the energy levels 2 → 1, 3 → 1, and 3 → 2 of an atom. Then**

- λ
_{21 }> λ_{31} - λ
_{21 }> λ_{32} - λ
_{31 }> λ_{32} - λ
_{31 }> λ_{32}

**Answer:** 1 and 3

**Question 41. Energy in the second energy state of a hydrogen atom is E _{2} Then**

- Energy in the third energy state of He
^{+}ion (Z =.2) is - Energy in the ground state of He
^{+}ion (Z = 2) is 16E_{2} - Energy in the third energy state of Li
^{2+}ion (Z = 3) is - Energy in the second energy state of Li2+ ion (Z = 3) is 9E
_{2}

**Answer:** 2, 3, and 4

**Question 42. The radius of the orbit of an electron in the ground state of a hydrogen atom is a _{0}. Then**

- The radius of the orbit of an electron in the second energy state of the He
^{+}ion (Z = 2) is 2a_{0} - The radius of orbit of an electron in the third energy state of He+ ion (Z = 2) is 3a
_{0} - The radius of orbit of an electron in the ground state of Li
^{2+}ion (Z = 3) is a_{0}/3 - The radius of orbit of an electron in the third energy state of Li
^{2+}ion (Z = 3) is 3a_{0}

**Answer:** 1, 3 and 4

**Question 43. In the hydrogen atom spectrum**

- Lines of the Balmer series are in the visible region
- Lines of the Lyman series are in the ultraviolet region
- Lines of the Paschen series are in the ultraviolet region
- Lines of Brackett series arc in the infrared region

**Answer:** 1, 2 and 3

**Question 44. In the case of X-ray spectrum**

- Cut-off wavelength depends on the kinetic energy of the electrons incident on the target
- Cut-off wavelength depends on. the material of the target
- Wavelength of Ka -line depends on the material of the target
- The Wavelength of Kg -line is larger than the wavelength of

**Answer:** 1 and 3

**Question 45. H, He ^{+}, and Li^{2+} are examples of .atoms or ions with one electron each. The energy of such atoms when in the n-th energy state (according to Bohr’s theory, n = 1, 2, 3,……. = principal quantum number) is E_{n} = \(-\frac{13.6 \mathrm{Z}^2}{n^2}\) (1eV = 1.6 × 10^{-19} J). For the ground state, n = 1 . To raise the atom from the ground state to n = f, the suitable incident light should have a wavelength given by λ = \(\frac{h c}{E_f-E_1}\) But the atom cannot stay permanently in the E_{f }– E_{1}, energy state, ultimately, it comes to the ground state by radiating extra energy, E_{f }– E_{1}, as electromagnetic radiation. The electron of the atom comes from n = f to n = 1 in one or more steps using the permitted energy levels. As a result, there is a possibility of emission of radia¬tion with more than one wavelength from the atom. Planck’s constant = 6.63 × 10^{-34} J.s. s and velocity of light c = 3× 10^{8 }m.s^{-1}**

**1. What Is the wavelength of the light incident on thr atom to raise it to the fourth quantum level hum ground stale?**

- 952 Å
- 975 Å
- 1027 Å
- 1219 Å

**Answer:** 2. 975 Å

**2. Radiations of how many wavelengths are possible in case of die excited atom In example 1 to come to ground state?**

- 2
- 3
- 6
- 9

**Answer: **3. 6

**3. What is the value of the maximum wavelength in example**

- 952 Å
- 975 Å
- 577 Å
- 10630 Å

**Answer: ** 4. 10630 Å

**4. What is the value of the minimum wavelength in example **

- 952 Å
- 975 Å
- 6577 Å
- 18830 Å

**Answer:** 2. 975 Å

**5. Energy of which quantum state of He ^{+} ion will be equal to the ground-level energy of hydrogen?**

- n = 1
- n = 2
- n = 3
- n = 4

**Answer:** 2. n= 2

**6. The wavelength of radiation emitted for the transition of the electron of He4 ion from n = 1 to n & 2 Is**

- 952 Å
- 975 Å
- 1027 Å
- 1219 Å

**Answer:** 4. 1219 Å

**7. For what wavelength of Incident radiation He ^{+} ion will be raised to the fourth quantum state from the ground state?**

- 243.7 Å
- 487.5 Å
- 731.2 Å
- 975 Å

**Answer:** 1. 243.7 Å

**8. Which among the following differences In the energy levels for a Li ^{2+} Ion is minimum?**

- E
_{2}– E_{1} - E
_{3}– E_{2} - E
_{3}– E_{1} - E
_{4}– E_{3}

**Answer: ** 4. E_{4} – E_{3}

**Question 46. The total energy of an electron for any particular energy level in a hydrogen atom Is -1,51 eV. The value of the principle quantum number of the energy level is**

- n = 2
- n = 1
- n = 3
- n = 4

**Answer:** 3. n = 3

The energy of electron In n -th energy level, wavelength

⇒ \(E_n=-\frac{13.6}{n^2}\)

-1. 51 = \(-\frac{13.6}{n^2}\)

Or, n = \(\sqrt{\frac{13.6}{1.51}}\)

= 3

**Question 47. The ratio of the minimum wavelength of Lyman and Balmer scries In the hydrogen spectrum will be**

- 10
- 5
- 0.25
- 1.25

**Answer:** 3. 0.25

**Question 48. If V is the accelerating voltage, then the maximum frequency of X-ray emitted from an X-ray tube is**

- \(\frac{e h}{V}\)
- \(\frac{e V}{h}\)
- \(\frac{h}{e V}\)
- None of these

**Answer:** 2. \(\frac{e V}{h}\)

The energy of an electron = eV = Iwmax = maximum energy of an X-ray photon

⇒ \(\nu_{\max }=\frac{e V}{h}\)

**Question 49. The ionization energy of hydrogen is 13.6 eV. The energy of tire photon released when an electron jumps from the first excited state (n = 2) to the ground state of a hydrogen atom is**

- 3.4 eV
- 4.53 eV
- 10.2 eV
- 13.6 eV

**Answer:** 3. 10.2 eV

The energy of the released photon

= \(13.6\left[\frac{1}{n_i^2}-\frac{1}{n_f^2}\right]\)

= \(13.6\left[\frac{1}{1^2}-\frac{1}{2^2}\right]\)

= 10.2 eV

**Question 50. A photon of wavelength 300 nm interacts with a stationary hydrogen atom in the ground state. During the interaction, the whole energy of the tire photon is transferred to the electron of the atom. State which possibility is correct considering Planck’s constant = 4 × 40 ^{-15}eV velocity of light = 3 × 10^{8} m/s, the ionization energy of hydrogen =13.6 eV.**

- The election will be knocked out of the atom
- Electrons will go to any excited seas of dead atom
- The electron will go only to the first excited state of the atom
- The electron will keep orbiting in the ground state of an atom

**Answer:** 4. Electron will keep orbiting in the ground state of

Energy ofthe photon

E = \(hf\frac{h c}{\lambda}=\frac{\left(4 \times 10^{-15}\right) \times\left(3 \times 10^6\right)}{300 \times 10^{-9}}\)

= 4 eV

Groundstate energy of hydrogen atom = – 13.6eV

Energy ofthe second orbit = \(\frac{13.6}{2^2}\)

= -3.4 eV

∴ The energy required to move the electron from the ground state second orbit = -3.4- (-13.6) = 10.2 eV.

Hence, the electron will keep orbiting the groundstate of the atom

**Question 51. The number of Broglie wavelengths contained in the second Bohr orbit of a Hydrogen atom is**

**Answer:**

- 1
- 2
- 3
- 4

**Answer:** 2. 2

**Question 52. The wavelength of the second Balmer line in the Hydrogen spectrum is 600 nm. The wavelength for its third line in the Lyntann series is**

**Answer:**

- 800
- 600
- 400
- None of these

**Answer: ** 4. None of these

We know, for the second line in the Balmer series.

⇒ \(\frac{1}{\lambda_1}=R\left[\frac{1}{4}-\frac{1}{16}\right]=\frac{3 R}{16}\)

For tliini line in Lyniann series,

⇒ \(\frac{1}{\lambda_2}=R\left[\frac{1}{1}-\frac{1}{16}\right]=\frac{15}{16} R\)

⇒ \(\frac{\lambda_2}{\lambda_1}=\frac{3 R}{16} \times \frac{16}{15 R}=\frac{1}{5}\)

Or, \(\lambda_2=\frac{1}{5} \times \lambda_1=\frac{1}{5} \times 600 \mathrm{~nm}\)

= 120 nm

**Question 53. Let vn and En be the respective speed and energy of an electron in the -th orbit of radius r _{n}, in a hydrogen atom, as predicted by Bohr’s model. Then**

- The plot of \(E_n r_n / E_1 r_1\) as a function of n is a straight line of slope 0
- The plot of \(r_n v_n / r_1 v_1\) as a function of n is a straight line of slope 1
- Plot of In \(\left(\frac{r_n}{r_1}\right)\) as a function of In(n) is a straight line of slope 2
- Plot of \(\left(\frac{r_n E_1}{E_n r_1}\right) \) as a function of In(n) is a straight line of slope 4

**Answer: ** 1,2, 3 and 4

According to Bohr’s theory, vn \(v_n \propto \frac{1}{n}\) ……………. (1)

\(E_n \propto \frac{1}{n^2}\) ……………. (2)

And \(r_n \propto n^2\) ……………. (3)

From equations (2) and (3) we get

⇒ \(E_n r_n \propto n^2 \times \frac{1}{n^2} \text { or, } E_n r_n \propto n^0\)

Hence En rn = constant

⇒ \(\frac{E_n r_n}{E_1 r_1}\)

From equations (1) and (3) we get, rnt/n« n

⇒ \(\frac{r_n v_n}{r_1 v_1}=n \text { or, } \frac{\left(r_n v_n\right) /\left(r_1 v_1\right)}{n}\)

= 1

From equation (3) \(\frac{r_n}{E_n} \propto n^4\)

⇒ \(\frac{r_n E_1}{E_n r_1}=n^4 \text { or, } \frac{\ln \left(r_n E_1 / E_n r_1\right)}{\ln (n)}\) = 4

**Question 54. How the linear velocity v of an electron in the Bohr orbit Is related to its quantum number n?**

- \(v \propto \frac{1}{n}\)
- \(v \propto \frac{1}{n^2}\)
- \(v \propto \frac{1}{\sqrt{n}}\)
- \(\nu \propto n\)

**Answer:** 1. \(v \propto \frac{1}{n}\)

We know, for n -th orbit

⇒ \(m v_n r_n=\frac{n h}{2 \pi}\)

Or, \(v_n=\frac{n h}{2 \pi m r_n}\)

⇒ \(v_n \propto \frac{1}{n}\left[r_n=\frac{\epsilon_0 n^2 h^2}{\pi m Z e^2}\right]\)

**Question 55. The radiation corresponding to the 3 → 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10 ^{-4} T.If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function ofthe metal is close to**

- 1.6 eV
- 1.8 eV
- 1.1 eV
- 0.8 eV

**Answer:** 3. 1.1 eV

The kinetic energy of an electron

= \(E_K=\frac{q^2 B^2 R^2}{2 m}\)

= \(=\frac{\left(1.6 \times 10^{-19}\right)^2 \times\left(3 \times 10^{-4}\right)^2 \times\left(10 \times 10^{-3}\right)^2}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}}\)

= 0.79 eV

For the transition of an electron from energy state 3 to 2

E = \(13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{13.6 \times 5}{36}\)

= 1.88 eV

According to Einstein’s photoelectric equation

E = E_{K}+ W_{0}

Or, W_{0} = E- E_{k}

= 1.88 – 0.79 ≈ 1.1 ev

**Question 56. Hydrogen (¹H _{1}), deuterium (²H_{1})), singly ionized helium (^{4}He_{2})^{+,} and doubly ionized lithium (^{6}Li_{3})^{++} all have electrons around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are, λ_{1}, λ_{2}, λ_{3 }and λ_{4} respectively then approximately which one of the following is correct?**

- λ
_{1 }= 2λ_{2}= 3λ_{3}= 4λ_{4} - 4λ
_{1}= 2λ_{2}=2λ_{2}= λ_{4} - λ
_{1}= 2λ_{2}= 2λ_{3}, = λ_{4} - λ
_{1}= λ_{2}= 4λ_{3}= 9λ_{4}

**Answer:** 4. λ_{1}= λ_{2}= 4λ_{3}= 9λ_{4}

For hydrogen, Z = 1 , for deuterium, Z = 1

For helium ion, Z = 2 ; for lithium-ion, Z = 3

**Question 57. As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom ion**

- Its kinetic energy increases but potential energy and total energy decrease
- Kinetic energy, potential energy, and total energy decrease
- Kinetic energy decreases, potential energy increases but total energy remains the same
- Kinetic energy and total energy decrease but potential

**Answer: ** 1. Its kinetic energy increases but potential energy and total energy decrease

The kinetic energy, potential energy, and total energy of a revolving electron in an atom or ion are E_{k}, E_{p}, and E respectively.

E_{K} + E_{p} = E-,E_{K} = -E-,E_{p}= 2E

∴ E_{K} is always positive,

∴ E_{p} and E are always negative

E gets decreased as an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ ion.

Hence, E_{p} also decreases, but E_{K} increases

**Question 58. An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If λ _{min} is the smallest possible wavelength of X-ray in the spectrum, the variation of log λ_{min }with IogV is correctly represented in**

**Answer: ** 1

⇒ \(\lambda_{\min }=\frac{h c}{e V}\)

Or, \(\log \lambda_{\min }=-\log V+\log \left(\frac{h c}{e}\right)\)

The negative sign implies a negative slope

**Question 59. Some energy levels of a molecule are. The ratio ofthe wavelengths r = λ _{1}/ λ_{2 }is given by**

- \(r=\frac{4}{3}\)
- \(r=\frac{2}{3}\)
- \(r=\frac{3}{4}\)
- \(r=\frac{1}{3}\)

**Answer:** 4. \(r=\frac{1}{3}\)

According to Bohr’s postulate

ΔE = hν = \(h \frac{c}{\lambda}\)

Or, \(\lambda=\frac{h c}{\Delta E}\)

In this case \(\frac{\lambda_1}{\lambda_2}=\frac{\Delta E_2}{\Delta E_1}=\frac{\left(-\frac{4}{3} E\right)-(-E)}{(-2 E)-(-E)}\)

⇒ \(\frac{1}{3}\)

**Question 60. If the series limit frequency of the Lyman series is vL, then the series limit frequency of the Pfund series is**

- \(\frac{\nu_L}{16}\)
- \(\frac{\nu_L}{25}\)
- 25ν
_{L} - 16ν
_{L}

**Answer:** 2. \(\frac{\nu_L}{25}\)

For the series limit frequency of the Lyman series

\(h \nu_L=E\left[\frac{1}{1^2}-\frac{1}{\infty}\right]\) = E

Again, for the series limit frequency of the Pfund series,

⇒ \(h \nu_p=E\left[\frac{1}{5^2}-\frac{1}{\infty}\right]=\frac{E}{25}\)

⇒ \(\nu_P=\frac{\nu_L}{25}\)

**Question 61. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let An, A& be the de Broglie wavelength of the electron in the n -th state and the groundstate respectively. Let An be the wavelength of the emitted photon in the transition from the n -th state to the ground state. For large n,[A, B are constants)**

- \(\Lambda_n^2 \approx A+B \lambda_n^2\)
- \(\Lambda_n^2 \approx \lambda\)
- \(\Lambda_n \approx A+\frac{B}{\lambda_n^2}\)
- \(\Lambda_n \approx A+B \lambda_n\)

**Answer:** 3. \(\Lambda_n \approx A+\frac{B}{\lambda_n^2}\)

We Know,

**Question 62. A hydrogen atom in the ground state is excited by a monochromatic radiation of = 975 Å. The number of spectral lines in the resulting spectrum emitted will be**

- 3
- 2
- 6
- 10

**Answer:** 3. 6

From Rydberg formula

⇒ \(\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)\)

⇒ \(\left(1-\frac{1}{n^2}\right)=\frac{1}{\lambda R}=\frac{1}{975 \times 10^{-8} \times 109706}\)

= 0.334

⇒ \(\frac{1}{n^2}=1-0.934\)

= 0.065

Or, n = 15.36

or, n = 3.91 ≈ 4

∴ Number of line spectra = ^{n}C_{2} = ^{4}C_{2}

= 6

**Question 63. Consider the 3rd orbit of He ^{+} (Helium), using the non-relativistic approach, the speed of an electron in this orbit will be [given k = 9 × 10^{9} constants, Z = 2 and h (Planck’s constant) = 6.6 × 10^{-34} J.s ]**

- 2.92 × 10
^{6}m/s - 0.73 × 10
^{6}m/s - 1.46 × 10
^{6}m/s - 3.0 × 10
^{8}m/s

**Answer:** 2. 0.73 × 10^{6} m/s

According to Bohr’s quantum condition

⇒ \(m v_n r_n=n \frac{h}{2 \pi} \text { or, } v_n=\frac{n h}{2 \pi m r_n}\)

Again \(r_n \propto \frac{n^2}{m Z} \text { or, } m r_n \propto \frac{n^2}{Z}\)

Hence \(v_n \propto n \frac{Z}{n^2} \quad \text { i.e., } v_n \propto \frac{Z}{n}\)

For ground state of hydrogen, Z = 1 and n = 1

⇒ \(v=\frac{c}{137}\)

For rhird orbit of he Z= 2

⇒ \(\frac{c}{137} \cdot \frac{2}{3}=\frac{3 \times 10^8}{137} \times \frac{2}{3}\)

**Question 64. Given the value of the Rydberg constant is 10 ^{6} m^{-1}, wave number of the last line of the Balmer series in the hydrogen spectrum will be**

- 0.53× 10
^{7}m^{-1} - 0.25 × 10
^{7}m^{-1} - 2.5 × 10
^{7}m^{-1} - 0.025 × 10
^{4}m^{-1}

**Answer:** 2. 0.25 × 10^{7} m^{-1}

⇒ \(\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)

The wave number ofthe last line ofthe Balmer series,

⇒ \(\bar{\nu}=\frac{1}{\lambda}=10^7\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)\)

= 0.25 × 10^{7} m^{-1}

**Question 65. If the longest wavelength in the ultraviolet region of the hydrogen spectrum is the shortest wavelength in its infrared region is**

- \(\frac{46}{7} \lambda_0\)
- \(\frac{20}{3} \lambda_0\)
- \(\frac{36}{5} \lambda_0\)
- \(\frac{27}{4} \lambda_0\)

**Answer:** 4. \(\frac{27}{4} \lambda_0\)

The longest wavelength in the ultraviolet region of

⇒ \(\frac{1}{\lambda_0}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\)

= \(\frac{3}{4}\) R

Or, \(\lambda_0=\frac{4}{3 R}\)

Again, if the shortest wavelength in the infrared region of the hydrogen spectrum = the shortest wavelength in the Paschen series = A

⇒ \(\frac{1}{\lambda}=R\left(\frac{1}{3^2}-\frac{1}{\infty^2}\right)=\frac{R}{9}\)

Or, \(\lambda=\frac{9}{R}=\frac{27}{4} \lambda_0\)

**Question 66. The ratio of kinetic energy to the total energy of an electron in a Bohr orbit ofthe hydrogen atom is**

- 2: -1
- 1: – 1
- 1:1
- 1:-2

**Answer:** 2. 1: – 1

The total energy and the kinetic energy of an electron in n nth Bohr radius is E_{n} and E_{k} respectively

E_{k} = – E_{n}

Or, \(\frac{E_K}{E_n}=1:(-1)\)