CBSE Class 12 Physics Modern Physics Objective Questions

CBSE Class 12 Physics Modern Physics Objective Questions. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

OBJECTIVES
CLASS 12 PHYSICS : MODERN PHYSICS

1. The ratio of momenta of an electron of mass m_e and an alpha particle of mas m_α which are accelerated from rest by a potential difference of 250 volt is...
(a) \( \frac{me}{mα} \)
(b) \( \sqrt{\frac{m_e}{m_\alpha}} \)
(c) \( \sqrt{\frac{2m_e}{m_\alpha}} \)
(d) \( \sqrt{\frac{m_e}{2m_\alpha}} \)

Answer: (d) \( \sqrt{\frac{m_e}{2m_\alpha}} \)

1. A particle of mass m and charge q accelerated through V gains kinetic energy qV, so its momentum is p = √(2mqV).
2. For the electron q = e; for the alpha particle q = 2e.
3. p_e/p_α = √(2m_eeV/(2m_α(2e)V)) = √(m_e/(2m_α)).
4. Hence, the correct option is (d).

Teacher's Note:
a) An alpha particle carries twice the electron's charge, which gains it twice the energy.
b) Momentum depends on both the mass and the energy gained.

2. A potential difference of 20 kV is applied to the anti-cathode of an X-ray tube to attract electrons. The maximum speed of electrons is...
(a) 8.4 × 10⁶ ms^-1
(b) 8.4 × 10⁷ ms^-1
(c) 5.95 × 10⁶ ms^-1
(d) 5.95 × 10⁷ ms^-1

Answer: (b) 8.4 × 10⁷ ms^-1

1. The electron gains kinetic energy eV = 1.6 × 10⁻¹⁹ × 2 × 10⁴ = 3.2 × 10⁻¹⁵ J.
2. (1/2)mv² = eV, so v = √(2eV/m) = √(2 × 3.2 × 10⁻¹⁵/9.1 × 10⁻³¹).
3. v = √(7.03 × 10¹⁵) = 8.4 × 10⁷ m/s.
4. Hence, the correct option is (b).

Teacher's Note:
a) Convert kilovolts to volts before substituting.
b) The speed is still below the speed of light, so the non-relativistic formula is acceptable here.

3. A proton and an alpha particle have kinetic energies in the ratio 16 : 1. Assuming that mass of alpha particle is equal to four times the mass of a proton, the ratio of de Broglie's wavelength of proton to that of alpha particle is...
(a) 1 : 4
(b) 1 : 2
(c) 2 : 1
(d) 4 : 1

Answer: (b) 1 : 2

1. de Broglie wavelength λ = h/p = h/√(2mK).
2. λ_p/λ_α = √(m_αK_α/(m_pK_p)).
3. m_α/m_p = 4 and K_α/K_p = 1/16, so the ratio = √(4/16) = 1/2.
4. Hence, the ratio is 1 : 2 and the correct option is (b).

Teacher's Note:
a) Wavelength is inversely proportional to the square root of mass times kinetic energy.
b) The proton has the larger kinetic energy but the smaller mass.

4. The radius of a nucleus with mass number 16 is 3.0 Fermi. The radius of another nucleus with mass number 128 is
(a) 6 Fermi
(b) 12 Fermi
(c) 18 Fermi
(d) 24 Fermi

Answer: (a) 6 Fermi

1. The nuclear radius R = R₀A^1/3, so R is proportional to A^1/3.
2. R'/R = (128/16)^1/3 = 8^1/3 = 2.
3. R' = 2 × 3.0 = 6 Fermi.
4. Hence, the correct option is (a).

Teacher's Note:
a) Eight times the mass number doubles the radius.
b) The nuclear density is the same for all nuclei.

5. The end product of decay of ₉₀Th²³² is ₈₂Pb²⁰⁸. The number of alpha and beta particles emitted are
(a) 4, 6
(b) 6, 4
(c) 6, 0
(d) 6, 3

Answer: (b) 6, 4

1. Each α-particle reduces the mass number by 4. Change in A = 232 - 208 = 24, so the number of α-particles = 24/4 = 6.
2. 6 α-particles reduce the atomic number by 12: 90 - 12 = 78. The final atomic number is 82.
3. Each β⁻ particle raises Z by 1, so the number of β-particles = 82 - 78 = 4.
4. Hence, 6 α and 4 β particles and the correct option is (b).

Teacher's Note:
a) Find the α-particles from the change in mass number, then the β-particles from the change in charge.
b) Beta decay does not change the mass number.

6. Two radioactive substances X and Y initially contain equal number of nuclei. X has a half life of 1 hour and Y has a half life of 2 hour. After two hours, the activity of X to that of Y is
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 4

Answer: (a) 1 : 1

1. Let the initial number of nuclei be N₀. The decay constants are λ_X = ln 2/1 h and λ_Y = ln 2/2 h, so λ_X = 2λ_Y.
2. After 2 hours, X has gone through 2 half-lives, so N_X = N₀/4. Y has gone through 1 half-life, so N_Y = N₀/2.
3. Activity A = λN: A_X = 2λ_Y × N₀/4 = λ_YN₀/2 and A_Y = λ_Y × N₀/2.
4. A_X : A_Y = 1 : 1.
5. Hence, the correct option is (a).

Teacher's Note:
a) The printed key gives (b), which is wrong; the correct value is 1 : 1 because the faster decay constant of X is exactly balanced by its smaller number of remaining nuclei.
b) Activity depends on both λ and the number of nuclei left.

7. The half life of radium is 1620 years and its atomic weight is 226 kg per kilomole. The number of atoms that will decay from its 1 gram sample per second will be
(a) 3.61 × 10¹⁰
(b) 3.61 × 10¹²
(c) 3.6 × 10¹⁴
(d) 3.61 × 10¹⁶

Answer: (a) 3.61 × 10¹⁰

1. Number of atoms in 1 g = (6.02 × 10²³/226) = 2.66 × 10²¹.
2. Decay constant λ = 0.693/(1620 × 3.16 × 10⁷ s) = 1.36 × 10⁻¹¹ s⁻¹.
3. Number of decays per second = λN = 1.36 × 10⁻¹¹ × 2.66 × 10²¹ = 3.61 × 10¹⁰.
4. Hence, the correct option is (a).

Teacher's Note:
a) Convert the half-life from years to seconds.
b) 1 curie = 3.7 × 10¹⁰ decays per second, which matches this activity of radium.

8. Masses of two isobars ₂₉Cu⁶⁴ and ₃₀Zn⁶⁴ are 63.9298 units and 63.9292 units respectively. It can be concluded from this data that
(a) both isobars are stable
(b) Zn⁶⁴ is radioactive decaying to Cu⁶⁴ through β^- -decay
(c) Cu⁶⁴ is radioactive decaying to Zn⁶⁴ through γ-decay
(d) Cu⁶⁴ is radioactive decaying to Zn⁶⁴ through β^--decay

Answer: (d) Cu⁶⁴ is radioactive decaying to Zn⁶⁴ through β^--decay

1. A nucleus can decay only to a nucleus of smaller mass. Cu-64 (63.9298) is heavier than Zn-64 (63.9292).
2. Zn has atomic number 30 and Cu has 29, so the mass number stays the same while the atomic number increases by 1.
3. This change is a β⁻ decay: Cu⁶⁴ → Zn⁶⁴ + e^- + antineutrino.
4. Hence, the correct option is (d).

Teacher's Note:
a) In β⁻ decay the mass number is unchanged and Z increases by one.
b) γ-decay changes neither A nor Z.

9. A sample contains 16 gm of a radioactive material, the half life of which is 2 days. After 32 days, the amount of radioactive material left in the sample is
(a) less than 1 mg
(b) \( \frac{1}{4} \) gm
(c) \( \frac{1}{2} \) gm
(d) 1 gm

Answer: (a) less than 1 mg

1. Number of half-lives = 32/2 = 16.
2. Amount left = 16 × (1/2)¹⁶ = 16/65536 g = 2.44 × 10⁻⁴ g.
3. This is 0.244 mg, which is less than 1 mg.
4. Hence, the correct option is (a).

Teacher's Note:
a) Use N = N₀(1/2)ⁿ where n is the number of half-lives.
b) Convert grams to milligrams to compare: 1 mg = 10⁻³ g.

10. One gm of a radioactive material having a half life of 2 years is kept in a store for a duration of 4 years. The fraction of material which has decayed is
(a) \( \frac{1}{8} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{3}{4} \)

Answer: (d) \( \frac{3}{4} \)

1. 4 years is 2 half-lives.
2. Fraction left = (1/2)² = 1/4.
3. Fraction decayed = 1 - 1/4 = 3/4.
4. Hence, the correct option is (d).

Teacher's Note:
a) Do not confuse the fraction left with the fraction decayed.

11. The mass of 2.0 curie of U²³⁴ radioisotope is
(a) 7.4 × 10^-10 gm
(b) 1.438 × 10^-11 gm
(c) 2.88 × 10^-11 gm
(d) 3.7 × 10^-10 gm

Answer: (c) 2.88 × 10^-11 gm

1. Activity = 2.0 curie = 2 × 3.7 × 10¹⁰ = 7.4 × 10¹⁰ decays per second.
2. Number of atoms N = activity/λ. The options correspond to a decay constant λ of 1 s⁻¹, so N = 7.4 × 10¹⁰.
3. Mass = N × 234/(6.02 × 10²³) = 7.4 × 10¹⁰ × 234/(6.02 × 10²³) = 2.88 × 10⁻¹¹ g.
4. Hence, the correct option is (c).

Teacher's Note:
a) No half-life is printed in the question; the printed options fit a decay constant of 1 s⁻¹. With the real half-life of U-234 (2.46 × 10⁵ years) the mass of 2 curie would be about 0.3 g.
b) Mass = (number of atoms × mass number)/Avogadro number.

12. A radioactive material has a mean life of 1620 years and 540 years for α and β emission respectively. The time in which one half of the material remains undecayed emitting both these radiations is
(a) 202.5 yrs
(b) 280.7 yrs
(c) 405 yrs
(d) 1080 yrs

Answer: (b) 280.7 yrs

1. The decay constants add: λ = λ_α + λ_β = 1/1620 + 1/540 = 4/1620 per year.
2. The mean life of the combined process = 1620/4 = 405 years.
3. Half-life = 0.693 × mean life = 0.693 × 405 = 280.7 years.
4. Hence, the correct option is (b).

Teacher's Note:
a) Decay constants add for parallel decay channels; the mean lives do not.
b) Half-life = 0.693 × mean life, not equal to the mean life.

13. Consider the spectral line resulting from the transition n = 2 → n = 1 in the atoms and ions given below. The shortest wavelength is produce by
(a) hydrogen atom
(b) deuterium atom
(c) singly ionized helium
(d) doubly ionized lithium

Answer: (d) doubly ionized lithium

1. For a hydrogen-like ion, 1/λ = RZ²(1/1² - 1/2²) = (3/4)RZ².
2. The wavelength is inversely proportional to Z². The largest Z gives the shortest wavelength.
3. Z = 1 for hydrogen and deuterium, Z = 2 for He⁺ and Z = 3 for Li²⁺.
4. Hence, the correct option is (d).

Teacher's Note:
a) Deuterium differs from hydrogen only slightly because of nuclear mass, not charge.
b) Energy levels scale as Z².

14. The ratio of the times taken by the electron in a hydrogen atom to complete one revolution in orbits corresponding to ground and first excited states is
(a) 1 : 4
(b) 4 : 1
(c) 1 : 8
(d) 8 : 1

Answer: (c) 1 : 8

1. In the Bohr model, the time period T = 2πr/v is proportional to n³ (r is proportional to n², v is proportional to 1/n).
2. Ground state n = 1 and first excited state n = 2.
3. T₁ : T₂ = 1³ : 2³ = 1 : 8.
4. Hence, the correct option is (c).

Teacher's Note:
a) The first excited state of hydrogen is n = 2.
b) The time period depends on the cube of n; the frequency is proportional to 1/n³.

15. The ratio of the longest and shortest wavelengths of the Lyman series is approximately
(a) 4/3
(b) 9/4
(c) 9/5
(d) 16/7

Answer: (a) 4/3

1. Lyman series: 1/λ = R(1 - 1/n²), n = 2, 3, 4, ...
2. Longest wavelength (n = 2): 1/λ_max = R(1 - 1/4) = 3R/4.
3. Shortest wavelength (n = ∞): 1/λ_min = R.
4. λ_max/λ_min = R/(3R/4) = 4/3.
5. Hence, the correct option is (a).

Teacher's Note:
a) The series limit corresponds to n = ∞.
b) The longest wavelength of any series comes from the transition between adjacent levels.

16. The longest wavelength of the Lyman series is approximately
(a) 920 Å
(b) 1020 Å
(c) 1220 Å
(d) 1420 Å

Answer: (c) 1220 Å

1. The longest wavelength of the Lyman series is the transition n = 2 to n = 1.
2. 1/λ = R(3/4) = 1.097 × 10⁷ × 0.75 = 8.23 × 10⁶ m⁻¹.
3. λ = 1.215 × 10⁻⁷ m = 1215 Å, which is about 1220 Å.
4. Hence, the correct option is (c).

Teacher's Note:
a) Using 12400 eV Å and ΔE = 10.2 eV gives the same wavelength, 1216 Å.
b) This line is in the ultraviolet region.

17. The first excitation potential of an atom is 10.2 V. Its ionization potential is
(a) 40.8 V
(b) 30.6 V
(c) 20.4 V
(d) 13.6 V

Answer: (d) 13.6 V

1. The atom is taken to be hydrogen-like. The first excitation energy from n = 1 to n = 2 is E₂ - E₁ = (3/4) × 13.6 = 10.2 eV, and the ionization energy is the full 13.6 eV.
2. Ionization potential = 10.2 × (4/3) = 13.6 V.
3. Hence, the correct option is (d).

Teacher's Note:
a) The first excitation energy is 3/4 of the ionization energy for a hydrogen-like atom.
b) An excitation potential of 10.2 V is the hydrogen value.

18. Energy levels A, B, C of a certain atom correspond to increasing values of energy, i.e., E_A < E_B < E_C. If λ₁, λ₁, λ₃ are the wavelength of radiation corresponding to the transitions C → B, B → A and C → A respectively, then
(a) λ₃ = λ₁ + λ₂
(b) λ₃ = \( \frac{λ1λ2}{λ1 + λ2} \)
(c) λ₃² = λ₁² + λ₂²
(d) λ₁ = λ₂ + λ₃

Answer: (b) λ₃ = \( \frac{λ1λ2}{λ1 + λ2} \)

1. The energy of the transition C → A is the sum of the energies of C → B and B → A: E_C - E_A = (E_C - E_B) + (E_B - E_A).
2. So hc/λ₃ = hc/λ₁ + hc/λ₂, that is, 1/λ₃ = 1/λ₁ + 1/λ₂.
3. λ₃ = λ₁λ₂/(λ₁ + λ₂).
4. Hence, the correct option is (b).

Teacher's Note:
a) The wavelengths do not add; the reciprocals of the wavelengths do.
b) The question lists λ₁ twice; the second one stands for λ₂ (B → A).

19. An electron jumps from the fourth to the second orbit of hydrogen atom. Given the Rydberg constant R = 10⁷ m^-1, the frequency of the emitted radiation, in Hz, is
(a) (3/16) 10¹⁵
(b) (3/4) 10¹⁵
(c) (9/4) 10¹⁵
(d) (9/16) 10¹⁵

Answer: (d) (9/16) 10¹⁵

1. 1/λ = R(1/2² - 1/4²) = 10⁷ × (1/4 - 1/16) = 10⁷ × 3/16 m⁻¹.
2. Frequency ν = c/λ = cR(3/16) = 3 × 10⁸ × 10⁷ × 3/16.
3. ν = (9/16) × 10¹⁵ Hz.
4. Hence, the correct option is (d).

Teacher's Note:
a) Use ν = c × (1/λ) after finding the wave number.
b) Keep the factor 3 of the speed of light, 3 × 10⁸ m/s.

20. The first spectral line of sodium is 5890 Å. The first excitation potential of sodium is
(a) 2.1 V
(b) 3.7 V
(c) 4.1 V
(d) 7.4 V

Answer: (a) 2.1 V

1. The first spectral line corresponds to the transition from the first excited state to the ground state.
2. Energy of the photon E = hc/λ = 12400/5890 eV = 2.1 eV.
3. The first excitation potential is therefore 2.1 V.
4. Hence, the correct option is (a).

Teacher's Note:
a) hc = 12400 eV Å is a handy constant.
b) The excitation potential in volts equals the excitation energy in electron-volts.

21. When the electron in a hydrogen atom jumps from the second orbit to the first orbit, the wavelength of the radiation emitted is λ. When the electron jumps from the third to the first orbit, the wavelength of the radiation emitted is
(a) \( \frac{9}{4} \) λ
(b) \( \frac{4}{9} \) λ
(c) \( \frac{27}{32} \) λ
(d) \( \frac{32}{27} \) λ

Answer: (c) \( \frac{27}{32} \) λ

1. For the transition n = 2 → 1: 1/λ = R(1 - 1/4) = 3R/4, so λ = 4/(3R).
2. For the transition n = 3 → 1: 1/λ' = R(1 - 1/9) = 8R/9, so λ' = 9/(8R).
3. λ'/λ = (9/8) × (3/4) = 27/32.
4. Hence, the correct option is (c).

Teacher's Note:
a) A larger energy jump gives a shorter wavelength, so the ratio must be less than 1.
b) Both transitions end on the ground state, so they belong to the Lyman series.

22. A hydrogen-like atom has one electron revolving around a stationary nucleus. The energy required to excite the electron from the second orbit to the third orbit is 47.2 eV. The atomic number of the atom is
(a) 3
(b) 4
(c) 5
(d) 6

Answer: (c) 5

1. Energy of the n-th level: E_n = -13.6 Z²/n² eV.
2. E₃ - E₂ = 13.6 Z²(1/4 - 1/9) = 13.6 Z² × 5/36 = 1.889 Z² eV.
3. 1.889 Z² = 47.2, so Z² = 25 and Z = 5.
4. Hence, the correct option is (c).

Teacher's Note:
a) Hydrogen-like energy levels are Z² times those of hydrogen.
b) Check: 1.889 × 25 = 47.2 eV.

23. The state of triply ionized beryllium, which has the same orbital radius as the ground state of hydrogen, has the principal quantum number
(a) 2
(b) 3
(c) 4
(d) 5

Answer: (a) 2

1. The orbital radius r_n = 0.53 n²/Z Å.
2. For hydrogen in the ground state, r = 0.53 Å (n = 1, Z = 1).
3. For Be³⁺, Z = 4: 0.53 n²/4 = 0.53, so n² = 4 and n = 2.
4. Hence, the correct option is (a).

Teacher's Note:
a) Triply ionized beryllium has Z = 4 and only one electron left.
b) The radius of the n-th orbit is proportional to n²/Z.

24. Hydrogen atoms in ground states are excited by monochromatic radiation of wavelength 960 Å. The number of lines in the resulting spectrum will be
(a) 3
(b) 4
(c) 6
(d) 10

Answer: (c) 6

1. Photon energy E = 12400/960 = 12.9 eV.
2. The energy of level n above the ground state is 13.6(1 - 1/n²): n = 4 needs 12.75 eV and n = 5 needs 13.06 eV.
3. The photon energy of 12.9 eV is not enough for n = 5, so the electron reaches at most n = 4.
4. Number of lines from n = 4 = 4 × 3/2 = 6.
5. Hence, the correct option is (c).

Teacher's Note:
a) Use the number of spectral lines n(n - 1)/2 for the highest level reached.
b) The photon is taken to excite the atom to n = 4 (12.75 eV, which is closest to 12.9 eV).

25. When a beam of accelerated electrons hits a target, a continuous X-ray spectrum is emitted from the target. Which of the following wavelengths is absent in the X-rays spectrum, if the X-ray tube is operating at 40,000 V ?
(a) 0.25 Å
(b) 0.5 Å
(c) 1.5 Å
(d) 1.0 Å

Answer: (a) 0.25 Å

1. The minimum wavelength of the continuous X-ray spectrum is λ_min = 12400/V Å = 12400/40000 = 0.31 Å.
2. No X-ray photons can have a wavelength shorter than 0.31 Å.
3. Only 0.25 Å is shorter than 0.31 Å, so it is absent.
4. Hence, the correct option is (a).

Teacher's Note:
a) The Duane-Hunt limit sets the shortest wavelength: eV = hc/λ_min.
b) All longer wavelengths are present in the continuous spectrum.

26. A hydrogen atom in its ground state absorbs 10.2 eV of energy. Its orbital angular momentum is increased by (given h = 6.6 × 10^-34 Js)
(a) 1.05 × 10^-34 Js
(b) 3.16 × 10^-34 Js
(c) 2.11 × 10^-34 Js
(d) 4.22 × 10^-34 Js

Answer: (a) 1.05 × 10^-34 Js

1. 10.2 eV is the energy needed for the transition from n = 1 to n = 2.
2. Angular momentum L = nh/2π. The change = (2 - 1)h/2π = h/2π.
3. h/2π = 6.6 × 10⁻³⁴/(2 × 3.14) = 1.05 × 10⁻³⁴ Js.
4. Hence, the correct option is (a).

Teacher's Note:
a) Angular momentum is quantised in units of h/2π.
b) 10.2 eV = (3/4) × 13.6 eV corresponds to n = 2.

27. The K_α X-ray emission line of tungsten occurs at λ = 0.021 nm. The energy difference between K and L levels in this atom is about
(a) 0.51 MeV
(b) 1.2 MeV
(c) 59 keV
(d) 13.6 eV

Answer: (c) 59 keV

1. The energy difference equals the energy of the emitted photon: E = hc/λ.
2. hc = 1240 eV nm, so E = 1240/0.021 = 5.9 × 10⁴ eV.
3. E = 59 keV.
4. Hence, the correct option is (c).

Teacher's Note:
a) hc = 1240 eV nm = 12400 eV Å is easy to remember.
b) X-ray energies are in the keV range.

28. Ionization energy of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. The number of spectral lines emitted by hydrogen atoms will be
(a) 1
(b) 2
(c) 3
(d) 6

Answer: (c) 3

1. The energy needed to excite the atom to level n is 13.6(1 - 1/n²). For n = 3: 13.6 × 8/9 = 12.09 eV, which matches 12.1 eV.
2. So the electron goes from n = 1 to n = 3.
3. Possible transitions on de-excitation: 3 → 2, 3 → 1 and 2 → 1, which gives 3 lines.
4. Hence, the correct option is (c).

Teacher's Note:
a) The number of lines is n(n - 1)/2 = 3 for n = 3.
b) The photon is absorbed only if its energy exactly equals an energy gap.