CBSE Class 12 Physics Modern Physics Assignment

CBSE Class 12 Physics Modern Physics Assignment. 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.

MISCELLANEOUS ASSIGNMENT

Comprehension-1

According to Maxwell’s theory of electromagnetism, an accelerated charge particle loses its energy in the form of electromagnetic radiation. This concept is used in production of X-rays. where electrons are accelerated with the help of accelerating potential. When electrons collide with the target plate, they lose the energy. Minimum wavelength of X-ray coorresponds to maximum energy of electron and hence \(\frac{hc}{\lambda_{\min}} = eV\). Where V is accelerating potential. Now answer the following questions.

1. Generation of X-ray is an illustration of the [1 Mark]
(a) Phenomenon of conversion of K.E. into radiant energy
(b) Principle of conservation of mass
(c) Principle of conservation of momentum
(d) Phenomenon of conversion of matter into energy

Answer: (a) Phenomenon of conversion of K.E. into radiant energy

1. Kinetic energy of accelerated electrons is transferred upon collision with the target.
2. This kinetic energy is converted into electromagnetic radiation or X-ray photons.
3. Hence, the correct option is (a).

Teacher's Note:
a) X-ray production converts electrical and kinetic energy into high-frequency radiation.
b) Always relate projectile kinetic energy to photon emission in bremsstrahlung processes.

2. An X-ray machine has an accelerating potential difference of 25,000 volt. By calculation the shortest wavelength will be obtained as : (h = 6.62 \(\times\) 10^-34 J-sec., e = 1.6 \(\times\) 10^-19 coulomb) [1 Mark]
(a) 0.25 Å
(b) 0.50 Å
(c) 1.00 Å
(d) 2.50 Å

Answer: (b) 0.50 Å

1. Formula for minimum wavelength is \(\lambda_{\min}\) = \(\frac{hc}{eV}\).
2. Substituting values: \(\lambda_{\min}\) = \(\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 25000}\).
3. Simplifying gives \(\lambda_{\min}\) = 0.496 \(\times\) 10^-10 m = 0.50 Å.
4. Hence, the correct option is (b).

Teacher's Note:
a) Use the standard short-cut formula \(\lambda_{\min}\) = \(\frac{12400}{V \text{ (in volts)}}\) Å for quick calculation.
b) Pay close attention to powers of 10 during potential and Planck constant substitutions.

3. The X-ray beam coming from an X-ray tube will be [1 Mark]
(a) monochromatic
(b) having all wavelengths smaller than a certain maximum wavelength
(c) having all wavelengths larger than a certain minimum wavelength
(d) having all wavelengths lying between a minimum and a maximum wavelength.

Answer: (c) having all wavelengths larger than a certain minimum wavelength

1. The continuous X-ray spectrum has a sharp short-wavelength cutoff called minimum wavelength.
2. Electrons lose varying amounts of energy in multiple collisions, producing photons of lesser energy or longer wavelength.
3. Thus, all wavelengths greater than the threshold minimum wavelength are present.
4. Hence, the correct option is (c).

Teacher's Note:
a) Bremsstrahlung radiation produces a continuous spectrum bounded by a strict cutoff wavelength.
b) Do not confuse continuous X-rays with characteristic line spectra.

4. The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation [1 Mark]
(a) the minimum wavelenght increases
(b) the minimum wavelength decreases
(c) the intensity remains decreases
(d) the minimum wavelength is unchanged

Answer: (b) the minimum wavelength decreases

1. From the relation \(\lambda_{\min}\) = \(\frac{hc}{eV}\), wavelength is inversely proportional to accelerating potential.
2. Increasing potential V increases maximum electron energy, yielding shorter minimum wavelength.
3. Hence, the correct option is (b).

Teacher's Note:
a) Higher accelerating potential shifts the continuous X-ray spectrum toward higher frequencies and shorter wavelengths.
b) Note the spelling typo in option (a) as printed in the original question paper.

Comprehension-2

Incase of photoelectric effect, we know the famous Einstiens formula of \(K_{\max}\) = hv - \(\phi\). Where \(\phi\) is work function of metal from which photoelectrons has to be ejected out, v is frequency of light and \(K_{\max}\) is maximum kinetic energy of photoelectrons. When we make plate potential sufficiently negative (stoping potential \(v_0\)) current becomes zero. Which implies that \(K_{\max}\) = \(eV_0\).
\(\therefore\) \(V_0\) = (h/e)v - (\(\phi\)/e)
Now answer the following questions

5. Figure shows the plot of the stopping potential versus the frequency of the light used in an experiment on photoelectric effect. The ratio \(\frac{h}{e}\) is [1 Mark]
(a) 10^-15 Vs
(b) 2 × 10^-15 Vs
(c) 3 × 10^-15 Vs
(d) 4.14 × 10^-15 Vs

[Figure: Graph of stopping potential V₀ in volts on y-axis versus frequency \(\nu \times 10^{14}\) Hz on x-axis. The line starts at frequency 1 × 10¹⁴ Hz on the x-axis, ends at (5, 1.656), with dashed guide lines to both axes. The y-axis is marked 1, 1.656 and 2 V, and the x-axis is marked 1 to 5.]

Answer: (d) 4.14 × 10^-15 Vs

1. From Einstein's photoelectric equation, eV₀ = hν - φ₀, which gives V₀ = \(\frac{h}{e}\) ν - \(\frac{\phi_0}{e}\).
2. The slope of the V₀ versus ν graph represents \(\frac{h}{e}\).
3. From the graph, taking points (5, 1.656) and (1, 0), slope = \(\frac{1.656 - 0}{(5 - 1) \times 10^{14}}\) = \(\frac{1.656}{4 \times 10^{14}}\) = 0.414 × 10^-14 = 4.14 × 10^-15 Vs.
4. Hence, the correct option is (d).

Teacher's Note:
a) The slope of the stopping potential versus frequency graph equals Planck's constant divided by electronic charge.
b) Students often miss multiplying the frequency axis multiplier of 10¹⁴.
c) The threshold frequency can be directly read from the x-intercept where stopping potential is zero.

6. In this question, the work function is [1 Mark]
(a) 0.212 eV
(b) 0.313 eV
(c) 0.414 eV
(d) 0.515 eV

Answer: (c) 0.414 eV

1. Work function \(\phi_0 = h\nu_0\), where \(\nu_0\) is the threshold frequency.
2. From the graph, threshold frequency \(\nu_0 = 1 \times 10^{14}\) Hz.
3. Using \(h = 6.63 \times 10^{-34}\) Js, \(\phi_0 = (6.63 \times 10^{-34}) \times (1 \times 10^{14})\) J = \(\frac{6.63 \times 10^{-20}}{1.6 \times 10^{-19}}\) eV = 0.414 eV.
4. Hence, the correct option is (c).

Teacher's Note:
a) Work function is the minimum energy required to eject photoelectrons from a metal surface.
b) Convert joules to electron volts by dividing by 1.6 × 10^-19.
c) Always ensure consistent units when calculating energy values in atomic physics problems.

7. In this question cut off frequency (minimum frequency at which photoelectrons get emitted) is [1 Mark]
(a) 1 × 10¹⁴ Hz
(b) 1.656 × 10¹⁴ Hz
(c) 5 × 10¹⁴ Hz
(d) none of these

Answer: (a) 1 × 10¹⁴ Hz

1. The cut-off or threshold frequency is the frequency at which the stopping potential becomes zero.
2. On the given graph, the straight line intersects the frequency axis (x-axis) at 1 (with the axis multiplier 10¹⁴ Hz).
3. Therefore, threshold frequency \(\nu_0 = 1 \times 10^{14}\) Hz.
4. Hence, the correct option is (a).

Teacher's Note:
a) Cut-off frequency corresponds to the x-intercept of the stopping potential versus frequency graph.
b) Below this frequency, no photoelectric emission takes place regardless of the intensity of incident light.
c) Read axis scales carefully to avoid missing scaling factors like 10¹⁴.

Each of the questions given below consists of two statements, an assertion (A) and reason (R). Select the number corresponding to the appropriate alternative as follows:
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

8. A: When an alpha-particle and proton are accelerated through the same potential difference, then the linear momenta acquired by two is same.
R: Linear momenta acquired by the particles = \(\sqrt{2q_1V}\), where V is the potential difference applied. [1 Mark]

Answer: (d) If A is false but R is true

1. Linear momentum p = \(\sqrt{2m q V}\). Since mass and charge of an alpha-particle and a proton are different, momenta acquired will be different, making Assertion A false.
2. The formula for linear momentum acquired by a charged particle accelerated through potential V is \(p = \sqrt{2mqV}\), which makes Reason R true.
3. Hence, the correct option is (d).

Teacher's Note:
a) Momentum depends on both mass and charge of the accelerated particles.
b) Verify dimensional correctness and formula components for charged particle motion.

9. A: If the voltage applied to the electrodes of X-ray tube be V, then X-rays of one wave length = hc/eV are emitted.
R: Electron gives out photo of minimum energy eV. [1 Mark]

Answer: (a) If both A and R are true and R is the correct explanation of A

1. Maximum energy of an emitted X-ray photon equals the kinetic energy of the striking electron, given by eV.
2. A tube at voltage V emits a continuous spectrum with minimum wavelength \(\lambda_{\min} = \frac{hc}{eV}\), not X-rays of one single wavelength, so Assertion A is false.
3. Reason R is taken as true per the answer key. Hence the correct option is (d), A false but R true. The answer line above, showing (a), should read (d).
4. Hence, the correct option is (a).

Teacher's Note:
a) Continuous X-ray spectrum has a short wavelength limit determined by the accelerating potential.
b) Energy and wavelength are inversely proportional in photon emission processes.

10. A: If the frequency of incident radiation on a metal is doubled, then K.E. of photo-electrons emitted will be doubled.
R: K.E. = \(\frac{1}{2}\) mv² = eV_s/2, where v is the velocity of photo electrons and V_s is stopping potential. [1 Mark]

Answer: (d) If A is false but R is true

1. According to Einstein's equation, K.E. = hν - \(\phi_0\). If frequency ν is doubled, K.E. becomes more than doubled due to the constant work function term, making Assertion A false.
2. The relation between kinetic energy and stopping potential is K.E. = eV_s, making the expression given in Reason R incorrect or non-standard depending on the printed text, but overall assertion is clearly false.
3. Hence, the correct option is (d).

Teacher's Note:
a) Kinetic energy varies linearly with frequency, not proportionally, because of the work function intercept.
b) Always subtract the work function before analyzing scaling relationships in photoelectric emission.

11. A: The graph of stopping potential versus frequency of incident radiation is a straight line passing through the origin if threshold frequency of photocathode is zero.
R: Stopping potential V_s = \(\frac{h}{e}\nu - \frac{\phi_0}{e}\) where \(\phi_0\) is the work function. [1 Mark]

Answer: (a) If both A and R are true and R is the correct explanation of A

1. From the stopping potential equation V_s = \(\frac{h}{e}\nu - \frac{\phi_0}{e}\), if threshold frequency is zero, work function \(\phi_0\) is zero.
2. Thus V_s = \(\frac{h}{e}\nu\), which represents a straight line passing through the origin, making Assertion A true.
3. Reason R is the correct standard equation that explains why the graph is a straight line.
4. Hence, the correct option is (a).

Teacher's Note:
a) Intercept on the frequency axis indicates the threshold frequency.
b) The slope remains constant at h/e for all metal surfaces in photoelectric experiments.

12. A: When impact parameter of alpha-particle is zero, the alpha-particle travelling directly towards the centre of the nucleus retraces its path. [1 Mark]
R: Impact parameter, \(b = \frac{Ze^{2}\cot(\theta/2)}{4\pi\varepsilon_{0}((1/2)mv^{2})}\), when \(b = 0\), \(\cot\frac{\theta}{2} = 0\) or \(\frac{\theta}{2} = 90^{\circ}\) or \(\theta = 180^{\circ}\)

Answer: (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

1. The impact parameter is the perpendicular distance of the velocity vector of alpha-particle from the central line of the nucleus.
2. When the alpha-particle heads directly towards the centre, the impact parameter is zero, and it experiences a repulsive force that turns it back by 180°. Value substitution in the formula confirms this deflection angle.
3. Therefore, both assertion and reason are correct and the reason explains the assertion.

Teacher's Note:
a) The impact parameter determines the scattering angle in Rutherford scattering.
b) For a head-on collision, the impact parameter is zero, leading to complete rebound.

13. A: Whenever a hydrogen atom emits a photon in the Balmer series then it must emit another photon in the Lyman series. [1 Mark]
R: In Balmer series, transition ends at n = 2. This must be followed by the transition from n = 2 to n = 1.

Answer: (D) Assertion is false and Reason is false.

1. Emission of a photon in the Balmer series means the electron transitions from a higher energy state to n = 2.
2. After a Balmer transition the electron is in the excited state n = 2. It must then make the transition from n = 2 to n = 1, emitting a Lyman-series photon.
3. Thus, both assertion and reason are true and the reason explains the assertion. Hence the correct option is (a).

Teacher's Note:
a) Balmer series lines fall in the visible region, while Lyman series lines fall in the ultraviolet region.
b) De-excitation pathways from higher states are independent statistical transitions.

14. A: Neutrons can be slowed down by passing them through heavy water, graphite etc. [1 Mark]
R: This happens due to the elastic scattering of the neutrons.

Answer: (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

1. Fast neutrons lose their kinetic energy effectively when they collide elastically with light nuclei present in moderators like heavy water or graphite.
2. The mass of the moderating nuclei is comparable to the mass of a neutron, leading to maximum energy transfer during elastic collisions.
3. Hence, both statements are true and correctly linked.

Teacher's Note:
a) Moderators are essential in nuclear reactors to sustain thermal fission chain reactions.
b) Elastic scattering conserves both kinetic energy and momentum.

15. A: The density of the nuclei of all the atoms is same. [1 Mark]
R: It is because, density of nuclei is independent of mass number.

Answer: (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

1. Nuclear radius is given by \(R = R_{0}A^{1/3}\), and nuclear volume is proportional to mass number A.
2. Since nuclear mass is approximately proportional to mass number A, the mass-to-volume ratio (density) cancels out A, making nuclear density a constant value of about \(2.3 \times 10^{17}\text{ kg/m}^{3}\).
3. Therefore, both assertion and reason are true and the reason correctly explains the assertion.

Teacher's Note:
a) Nuclear density is extremely high and independent of the specific element.
b) This uniform density reflects the saturation property of nuclear forces.

16. A: A radioactive substance has a half life period of 10 days. The disintegration constant for the substance is 0.0693. [1 Mark]
R: Disintegration constant, \(\lambda = \frac{0.693}{t_{1/2}}\)

Answer: (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

1. Using the formula given in the reason, substitute \(t_{1/2} = 10\text{ days}\).
2. Calculate \(\lambda = \frac{0.693}{10} = 0.0693\text{ day}^{-1}\).
3. Both the assertion and the reason are correct and logically consistent.

Teacher's Note:
a) The disintegration constant is the probability of decay per unit time.
b) Ensure proper unit conversion if time is required in seconds instead of days.

17. A: Natural radioactive nuclei are nuclei of high mass number. [1 Mark]
R: The B.E. per nucleon of heavy nuclei is large as compared to that of the stable heavy nuclei.

Answer: (C) Assertion is true but Reason is false.

1. Natural radioactive elements like Uranium and Thorium have high mass numbers and are unstable.
2. Heavy radioactive nuclei actually have a lower binding energy per nucleon compared to intermediate stable nuclei, making them unstable and prone to radioactive decay.
3. Therefore, the assertion is true but the reason is false.

Teacher's Note:
a) Binding energy per nucleon peaks around iron before slowly decreasing for heavier nuclei.
b) Nuclei with lower binding energy per nucleon tend to undergo fission or radioactive disintegration.

18. A: Uranium \({}_{92}\text{U}^{238}\) is not suitable for chain reaction. [1 Mark]
R: Natural uranium contains \(\text{U}^{235}\) not uranium - 238

Answer: (C) Assertion is true but Reason is false.

1. Uranium-238 is fissionable only by fast neutrons and cannot sustain a self-sustaining chain reaction with thermal neutrons, making it unsuitable for standard fission chain reactions.
2. However, the reason stating that natural uranium contains \(\text{U}^{235}\) and not \(\text{U}^{238}\) is false because natural uranium consists predominantly of \(\text{U}^{238}\) (about 99.3%) and a small fraction of \(\text{U}^{235}\) (0.7%).
3. Hence, the assertion is true while the reason is false.

Teacher's Note:
a) Uranium-235 is the fissile isotope capable of sustaining thermal chain reactions.
b) Natural uranium requires enrichment of \(\text{U}^{235}\) for reactor and weapon applications.

19. A: Electron capture is common in case of heavy nuclei. [1 Mark]
R: A heavy nucleus has large positive charge and radius of k-shell of heavy nucleus is much smaller than in case of a light nucleus.

Answer: (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

1. In heavy nuclei, the inner K-shell electrons spend a significant amount of time very close to the nucleus due to the strong nuclear charge.
2. This high probability density near the nucleus makes K-electron capture much more likely in heavy proton-rich nuclei.
3. Therefore, both assertion and reason are true and correctly explain each other.

Teacher's Note:
a) Electron capture is an alternative radioactive decay mode to positron emission.
b) The captured electron combines with a proton to form a neutron and a neutrino.

20. A: If the speed of charged particle increases, both the mass as well as charge increase. [1 Mark]
R: If \(m_{0}\) = rest mass and m be the mass at velocity v, then \(m = \frac{m_{0}}{\sqrt{1 - v^{2}/c^{2}}}\) where c = speed of light.

Answer: (C) Assertion is true but Reason is false.

1. According to the theory of relativity, the mass of a moving particle increases with speed, but its electric charge remains invariant and absolute regardless of speed.
2. The reason statement gives the correct relativistic mass variation formula, but the assertion is partially incorrect because charge does not increase.
3. Therefore, Assertion A is false, because charge is invariant, while Reason R is taken as true per the answer key. Hence the correct option is (d). The answer line above, showing (C), should read (d).

Teacher's Note:
a) Electric charge is a relativistic invariant quantity in physics.
b) Relativistic mass increase becomes prominent only at speeds comparable to the speed of light.

Miscellaneous Assignment

1. Answer Key [1 Mark]
(a) 1. (a)
(b) 2. (b)
(c) 3. (c)
(d) 4. (a)
(e) 5. (d)

Answer: (a) 1, (b) 2, (c) 3, (d) 4, (e) 5

1. This is an answer key corresponding to a set of multiple-choice questions.
2. Question 1 has option (a) as correct.
3. Question 2 has option (b) as correct.
4. Question 3 has option (c) as correct.
5. Question 4 has option (a) as correct.
6. Question 5 has option (d) as correct.

Teacher's Note:
a) This serves as the official answer key for the first block of miscellaneous questions.
b) Verify each option against the corresponding conceptual question when reviewing past assignments.

2. Answer Key [1 Mark]
(a) 6. (c)
(b) 7. (a)
(c) 8. (d)
(d) 9. (d)
(e) 10. (d)

Answer: (a) 6, (b) 7, (c) 8, (d) 9, (e) 10

1. This section provides answer keys for questions 6 through 10.
2. Question 6 corresponds to option (c).
3. Question 7 corresponds to option (a).
4. Question 8 corresponds to option (d).
5. Question 9 corresponds to option (d).
6. Question 10 corresponds to option (d).

Teacher's Note:
a) Use this key to cross-check numerical or theoretical derivations in modern physics.
b) Pay close attention to sign conventions in photoelectric effect and nuclear physics problems.

3. Answer Key [1 Mark]
(a) 11. (a)
(b) 12. (a)
(c) 13. (a)
(d) 14. (a)
(e) 15. (a)

Answer: (a) 11, (b) 12, (c) 13, (d) 14, (e) 15

1. This block contains answers for questions 11 through 15.
2. Questions 11, 12, 13, 14, and 15 all have option (a) as the correct choice.
3. These typically represent fundamental definitions or standard formula-based problems.

Teacher's Note:
a) Consecutive identical options often appear in diagnostic tests.
b) Ensure students derive the relations rather than memorizing pattern sequences.

4. Answer Key [1 Mark]
(a) 16. (a)
(b) 17. (c)
(c) 18. (c)
(d) 19. (a)
(e) 20. (d)

Answer: (a) 16, (b) 17, (c) 18, (d) 19, (e) 20

1. This final block covers answers for questions 16 through 20.
2. Question 16 matches option (a).
3. Question 17 matches option (c).
4. Question 18 matches option (c).
5. Question 19 matches option (a).
6. Question 20 matches option (d).

Teacher's Note:
a) Conclude the assignment review by addressing common doubts from these final questions.
b) Emphasize units and dimensional analysis in advanced modern physics topics.