CBSE Class 12 Physics Dual Nature Nucleus and Atom Assignment

TEST SERIES (PHYSICS)

DUAL NATURE,NUCLEUS AND ATOM 

Question. An electron of mass m and charge e initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t ignoring relativistic effects is
(a) -h/eEt²
(b) -eht/E
(c) -mh/eEt²
(d) -h/eE

Answer: A

Question. Radiations of intensity 0.5 W/m² are striking a metal plate.
The pressure on the plate is
(a) 0.166 × 10^–8 N/m²
(b) 0.332 × 10^–8 N/m²
(c) 0.111 × 10^–8 N/m²
(d) 0.083 × 10^–8 N/m²

Answer: A

Question. The magnitude of the de-Broglie wavelength (l) of electron (e), proton (p), neutron (n) and a-particle (a) all having the same energy of 1 MeV, in the increasing order will follow the sequence
(a) λ_e, λ_p, λ_n, λ_α
(b) λ_e, λ_n, λ_p, λ_α
(c) λ_α, λ_n, λ_p, λ_e
(d) λ_p, λ_e, λ_α, λn 

Answer: C

Question. Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell, the work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero the voltage of the anode relative to the cathode must be made
(a) – 4.2 V
(b) – 9.4 V
(c) – 17.8 V
(d) + 9.4 V

Answer: B

Question. Einstein’s work on photoelectric effect provided support for the equation
(a) E = hn
(b) E = mc²
(c) E = -Rhc/n²
(d) K.E. = 1/2 mv2

Answer: A

Question. The glancing angle in a X-ray diffraction is 30º and the wavelength of X-rays used is 20 nm. The interplanar spacing of the crystal dffracting these X-rays will be
(a) 40 nm
(b) 20 nm
(c) 15 nm
(d) 10 nm

Answer: B

Question. The maximum velocity of an electron emitted by light of wavelength λ incident on the surface of a metal of work function Φ is 

Answer: C

Question. The maximum distance between interatomic lattice planes is 15 Å. The maximum wavelength of X-rays which are diffracted by this crystal will be
(a) 15 Å
(b) 20 Å
(c) 30 Å
(d) 45 Å

Answer: C

Question. When a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V/4 . The threshold wavelength for the metallic surface is : 
(a) 4λ
(b) 5λ
(c) 5/2 λ
(d) 3λ

Answer: D

Question. If the X-ray tube is working at 20 kV then the minimum wavelength of X-rays will be
(a) 0.31 Å
(b) 0.62 Å
(c) 0.93 Å
(d) 0.47 Å

Answer: B

Question. A radio transmitter operates at a freqency 880 kHz and a power of 10 kW. The number of photons emitted per second is
(a) 1.72 × 10³¹
(b) 1.327 × 10²⁵
(c) 1.327 × 10³⁷
(d) 1.327 × 10⁴⁵

Answer: A

Question. For photoelectric emission from certain metal the cut-off frequency is n. If radiation of frequency 2n impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass) 
(a) √hn / m
(b) √2hn / m
(c) √2 hn / m
(d) √hn / (2m)

Answer: B

Question. The momentum of a photon of an electromagnetic radiation is 3.3 × 10^–29 kgms^–1. What is the frequency of the associated waves ?
[h = 6.6 × 10^–34 Js; c = 3 × 10⁸ ms^–1)
(a) 1.5 × 10¹³ Hz
(b) 7.5 × 10¹² Hz
(c) 6.0 × 10³ Hz
(d) 3.0 × 10³ Hz

Answer: A

Question. An electron is moving with an initial velocity v = v₀ î and is in a magnetic field B = B₀ ĵ . Then, it's de-Broglie wavelength 
(a) remains constant
(b) increases with time
(c) decreases with me
(d) increases and decreases periodically

Answer: A

Question. The wavelength of a 1 keV photon is 1.24 × 10^–9 m. What is the frequency of 1 MeV photon ?
(a) 1.24 × 10¹⁵
(b) 2.4 × 10²⁰
(c) 1.24 × 10¹⁸
(d) 2 × 4 × 10²³

Answer: B

Question. In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is
(a) 1.8 V
(b) 1.2 V
(c) 0.5 V
(d) 2.3 V

Answer: C

Question. A proton, a neutron, an electron and an a-particle have same energy. Then, their de-Broglie wavelengths compare as
(a) λ_p = λ_n > λ_e > λ_α
(b) λ_α < λ_p = λ_n > λ_e
(c) λ_e < λ_p = λ_n > λ_α
(d) λ_e = λ_p = λ_n = λα

Answer: B

Question. The frequency and work function of an incident photon are v and Φ0. If v₀ is the threshold frequency then necessary condition for the emission of photoelectron is
(a) v < v₀
(b) v = v₀/2
(c) v ≥ v₀
(d) None of these

Answer: C

Question. The wavelength le of an electron and lp of a photon are of same energy E are related by
(a) λ_p ∝ λ_e
(b) λ_p ∝ √λ_e
(c) λ_p ∝ 1/√λ_e
(d) λ_p ∝ λe²

Answer: D

Question. A source of light is placed at a distance of 50 cm from a photocell and the stopping potential is found to be V₀. If the distance between the light source and photocell is made 25 cm, the new stopping potential will be
(a) 2V₀
(b) V₀/2 
(c) V₀
(d) 4V0

Answer: C

Question. The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nuclear with 1 MeV energy is nearly
(a) 1.2 nm
(b) 1.2 × 10^–3 nm
(c) 1.2 × 10^–6 nm
(d) 1.2 × 10 nm

Answer: B

Question. The quantity e²/2hε₀c has a value
(a) 1/137 ms^-1
(b) 2/137 ms^-1
(c) 1/137
(d) 2/137

Answer: C

Question. The threshold frequency for a photosensitive metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly
(a) 2 V
(b) 3 V
(c) 5 V
(d) 1 V

Answer: A

Question. The de-Broglie wavelength of neutron in thermal equilibrium at temperature T is
(a) 30.8/√T Å
(b) 3.08/√T Å
(c) 0.308/√T Å
(d) 0.0308/√T Å

Answer: A

Question. When the energy of the incident radiation is incredased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is : 
(a) 0.65 eV
(b) 1.0 eV
(c) 1.3 eV
(d) 1.5 eV

Answer: B

Question. If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is : 
(a) 25
(b) 75
(c) 60
(d) 50

Answer: B

Question. Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is: 
(a) < 2.8 × 10^-9 m
(b) ≥. 2.8 × 10^-9 m
(c) ≤ 2.8 × 10^-12 m
(d) < 2.8 × 10^-10 m

Answer: B

Question. Consider a beam of electrons (each electron with energy E₀) incident on a metal surface kept in an evacuated chamber. Then,
(a) no electrons will be emitted as only photons can emit electrons
(b) electrons can be emitted but all with an energy, E₀
(c) electrons can be emitted with any energy, with a maximum of E₀ – Φ (Φ is the work function)
(d) electrons can be emitted with any energy, with a maximum of E0

Answer: D

Question. Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the materials is
(a) 4 × 10¹⁵ Hz
(b) 5 × 10¹⁵ Hz
(c) 1.6 × 10¹⁵ Hz
(d) 2.5 × 10¹⁵ Hz

Answer: C

Question. Consider figure given below. Suppose the voltage applied to A is increased. The diffracted beam will have the maximum at value of θ that 

(a) will be larger than the earlier value
(b) will be the same as the earlier value
(c) will be less than the earlier value
(d) will depend on the target

Answer: C
 

Q1. A proton and an alpha particle both initially at rest are accelerated so as to have the same kinetic energy. What is the ratio of their De Broglie wave length.
Answer: The de Broglie wavelength is given by the formula:
\(\lambda = \frac{h}{\sqrt{2mK}}\)
Since both particles have the same kinetic energy \(K\) and Planck's constant \(h\) is constant, we have:
\(\lambda \propto \frac{1}{\sqrt{m}}\)
Let mass of proton be \(m_p = m\), then mass of alpha particle is \(m_\alpha = 4m\).
Ratio of wavelengths is:
\(\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha}{m_p}} = \sqrt{\frac{4m}{m}} = \sqrt{4} = 2\)
Therefore, the required ratio is \(2:1\).
In simple words: The de Broglie wavelength is inversely proportional to the square root of the particle's mass when kinetic energy is constant.

Exam Tip: Remember that an alpha particle consists of 2 protons and 2 neutrons, so its mass is approximately 4 times that of a proton.

Q2. Maximum velocity of electron emitted from a metal of negligible work function is v when a radiation of frequency f falls on it. what would be the velocity of electrons when a radiation of frequency 4f falls on same metal surface.
Answer: According to Einstein's photoelectric equation for negligible work function (\(\phi_0 \approx 0\)):
\(\frac{1}{2}mv^2 = hf \implies v \propto \sqrt{f}\)
When frequency becomes \(4f\), let the new velocity be \(v'\):
\(\frac{1}{2}mv'^2 = h(4f)\)
Dividing the two equations:
\(\frac{v'}{v} = \sqrt{\frac{4f}{f}} = \sqrt{4} = 2\)
\(v' = 2v\)
In simple words: When the frequency of incident radiation is quadrupled, the maximum velocity of emitted electrons is doubled.

Exam Tip: Always state Einstein's photoelectric equation clearly before substituting the work function as zero.

Q3. Find the ratio of densities of two nuclii of which mass number are 27 and 64.
Answer: Nuclear density is independent of the mass number \(A\) because nuclear radius is given by \(R = R_0 A^{1/3}\) and nuclear volume is proportional to \(A\).
Mass of nucleus is proportional to \(A\).
\(\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{m A}{\frac{4}{3}\pi R_0^3 A} = \frac{3m}{4\pi R_0^3}\)
Since all terms in the expression are constant, the nuclear density is a constant value for all nuclei.
Therefore, the ratio of densities of the two nuclei is \(1:1\).
In simple words: Nuclear matter has a nearly constant density regardless of its mass number.

Exam Tip: This is a conceptual question frequently asked in board exams; always highlight that nuclear density is independent of mass number \(A\).

Q4. Calculate the half life of a radioactive substance of which the activity drops to \(1/16^{\text{th}}\) of its initial value in 30 years.
Answer: The relation between initial activity \(R_0\) and final activity \(R\) after time \(t\) is given by:
\(R = R_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}\)
Given that \(R = \frac{1}{16} R_0\) and \(t = 30\) years:
\(\frac{1}{16} R_0 = R_0 \left(\frac{1}{2}\right)^{30/T_{1/2}}\)
\(\left(\frac{1}{2}\right)^4 = \left(\frac{1}{2}\right)^{30/T_{1/2}}\)
Comparing the powers:
\(4 = \frac{30}{T_{1/2}}\)
\(T_{1/2} = \frac{30}{4} = 7.5 \text{ years}\)
In simple words: Since activity drops to \(1/16\) (which is \((1/2)^4\)) in 30 years, it takes 4 half-lives to reach this value, making each half-life 7.5 years.

Exam Tip: Express fractions as powers of 2 to quickly determine the number of half-lives elapsed.

Q5. For which transition shortest wavelength of Balmer series belongs?
Answer: The wavelength of spectral lines in the Balmer series is given by:
\(\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \text{ where } n = 3, 4, 5, \dots\)
For the shortest wavelength, the energy difference must be maximum, which occurs when the electron transitions from infinity (\(n = \infty\)) to the second orbit (\(n = 2\)).
Therefore, the transition is from \(n = \infty\) to \(n = 2\).
In simple words: The shortest wavelength corresponds to the highest energy transition, which is from the outermost orbit (\(\infty\)) to \(n = 2\).

Exam Tip: Shortest wavelength always corresponds to the series limit (highest possible upper level, \(n = \infty\)).

Q6. Write the points which indicates that photo electric effect proves to particle nature of light.
Answer: The points indicating that the photoelectric effect proves the particle nature of light are:
1. Threshold frequency: Emission of electrons depends only on the frequency of incident radiation, not on its intensity, which is explained by assuming light consists of discrete packets of energy called photons.
2. Instantaneous emission: Photoelectric emission is an instantaneous process with no time lag between the incidence of light and emission of electrons, supporting the collision-like particle nature of photons.
3. Intensity dependence: Increasing the intensity increases the number of emitted electrons (current) but does not change their maximum kinetic energy, proving that energy is transferred in localized packets (quanta).
In simple words: Wave theory fails to explain why light frequency dictates electron emission, whereas treating light as particle packets (photons) successfully explains it.

Exam Tip: Mention threshold frequency and instantaneous emission as primary arguments for the particle nature of light.

Q7. A radiation of wavelength \(\lambda\) falls on a metal surface of negligible work function. Find the deBroglie wavelength associated with emitted electrons.
Answer: Energy of the incident photon is given by:
\(E = \frac{hc}{\lambda}\)
Since work function is negligible, the entire photon energy is converted into kinetic energy \(K\) of the emitted electron:
\(K = \frac{hc}{\lambda}\)
The de Broglie wavelength \(\lambda_e\) of the emitted electron of mass \(m\) is:
\(\lambda_e = \frac{h}{\sqrt{2mK}}\)
Substituting \(K\):
\(\lambda_e = \frac{h}{\sqrt{2m \left(\frac{hc}{\lambda}\right)}} = \sqrt{\frac{h \lambda}{2mc}}\)
In simple words: We equate photon energy to electron kinetic energy and substitute it into the de Broglie wavelength formula.

Exam Tip: Keep track of symbols carefully; distinguish between the wavelength of incident radiation (\(\lambda\)) and the de Broglie wavelength of the electron (\(\lambda_e\)).

Q8. Draw a graph showing the variation of potential energy between a pair of nucleons as a function of their separation. Indicate the regions in which the nuclear force is (i) attractive (ii) repulsive
Answer:
Construction steps for the potential energy curve:
1. Draw the horizontal axis representing separation distance \(r\) (in fm) and the vertical axis representing potential energy \(V\) (in MeV).
2. Plot a deep potential well with a minimum at approximately \(r = 0.8 \text{ fm}\).
3. For separations \(r < 0.8 \text{ fm}\), the potential energy curve rises steeply, indicating the repulsive region of the nuclear force.
4. For separations between \(0.8 \text{ fm}\) and \(4 \text{ fm}\), the curve lies in the negative potential energy region with a minimum at \(0.8 \text{ fm}\), indicating the attractive region.

In simple words: The graph shows a potential well where nucleons repel at very short distances and attract at distances around 0.8 fm to a few fm.

Exam Tip: Label the minimum point at 0.8 fm clearly as it is a crucial standard value in nuclear physics.

Q9. Why is the mass of a nucleus always less than the sum of masses of its constituents, neutrons and protons. If the total number of neutrons and protons in a nuclear reaction is conserved, how then the energy absorbed or evolved in the reaction?
Answer: 1. The mass of a nucleus is less than the sum of masses of its constituent nucleons because a certain amount of mass is converted into binding energy when nucleons bind together to form a stable nucleus (mass defect, \(\Delta m = \sum m_{\text{nucleons}} - m_{\text{nucleus}}\)).
2. Although the total number of nucleons is conserved in a nuclear reaction, the total rest mass of the products is slightly different from the total rest mass of the reactants. This mass difference \(\Delta m\) is released or absorbed as energy according to Einstein's mass-energy equivalence relation \(E = \Delta m c^2\).
In simple words: Mass is converted into binding energy when nucleons join, and changes in total mass during reactions account for energy release or absorption.

Exam Tip: Mention Einstein's equation \(E = \Delta m c^2\) explicitly to connect mass defect with reaction energy.

Q10. By using the concept of deBroglie wavelength prove the Bohrs condition of quantization condition of angular momentum of an electron in an orbit.
Answer: According to de Broglie's hypothesis, an electron of momentum \(p = mv\) revolving in an orbit of radius \(r\) has an associated wavelength:
\(\lambda = \frac{h}{mv}\)
For an electron wave to form a stable standing wave in an orbit, the circumference of the orbit must be an integral multiple of its de Broglie wavelength:
\(2\pi r = n\lambda \text{ where } n = 1, 2, 3, \dots\)
Substituting the value of \(\lambda\):
\(2\pi r = n \left(\frac{h}{mv}\right)\)
Rearranging the terms to isolate angular momentum \(mvr\):
\(mvr = \frac{nh}{2\pi}\)
This is Bohr's quantization condition of angular momentum.
In simple words: Standing waves can only fit in orbits whose circumferences are whole number multiples of the de Broglie wavelength.

Exam Tip: Clearly state the standing wave condition \(2\pi r = n\lambda\) as the starting point of this derivation.

Q11. Derive an expression for the total energy of an electron in any orbit of hydrogen atom according to Bohr's model. what the negative sign signify?
Answer: For an electron of mass \(m\) revolving around a nucleus of charge \(+Ze\) in an orbit of radius \(r\) with velocity \(v\):
The electrostatic force provides the necessary centripetal force:
\(\frac{1}{4\pi \varepsilon_0} \frac{e^2}{r^2} = \frac{mv^2}{r} \implies mv^2 = \frac{1}{4\pi \varepsilon_0} \frac{e^2}{r}\)
Kinetic energy (\(K\)) is:
\(K = \frac{1}{2}mv^2 = \frac{1}{8\pi \varepsilon_0} \frac{e^2}{r}\)
Potential energy (\(U\)) is:
\(U = \frac{1}{4\pi \varepsilon_0} \frac{(+e)(-e)}{r} = -\frac{1}{4\pi \varepsilon_0} \frac{e^2}{r}\)
Total energy (\(E\)) is the sum of kinetic and potential energy:
\(E = K + U = \frac{1}{8\pi \varepsilon_0} \frac{e^2}{r} - \frac{1}{4\pi \varepsilon_0} \frac{e^2}{r} = -\frac{1}{8\pi \varepsilon_0} \frac{e^2}{r}\)
Significance of the negative sign: The negative sign signifies that the electron is bound to the nucleus by attractive electrostatic forces and cannot escape without external energy input.
In simple words: Total energy is negative because the attractive force keeps the electron bound in the atomic system.

Exam Tip: Remember the key relation \(U = -2K\) and \(E = -K\), which simplifies energy calculations in atomic physics.

Q12. Find the kinetic energy of electron in second excited state of hydrogen atom? If the electron jump to ground state from second excited state, find the wavelength of the spectral line emitted? (3)
Answer: 1. The principal quantum number for the second excited state is \(n = 3\).
Total energy in the \(n^{\text{th}}\) orbit of a hydrogen atom is given by:
\(E_n = -\frac{13.6}{n^2} \text{ eV}\)
For \(n = 3\):
\(E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.51 \text{ eV}\)
Since Kinetic Energy \(K = -E\):
\(K_3 = 1.51 \text{ eV}\)
2. When the electron jumps from the second excited state (\(n_2 = 3\)) to the ground state (\(n_1 = 1\)), the energy of the emitted photon is:
\(\Delta E = E_3 - E_1 = -1.51 \text{ eV} - (-13.6 \text{ eV}) = 12.09 \text{ eV}\)
The wavelength \(\lambda\) of the emitted spectral line is calculated as:
\(\lambda = \frac{hc}{\Delta E} = \frac{1242 \text{ eV}\cdot\text{nm}}{12.09 \text{ eV}} \approx 102.7 \text{ nm}\)
In simple words: We first find the energy at \(n = 3\), use its negative value for kinetic energy, and then calculate the wavelength from the energy difference between \(n = 3\) and \(n = 1\).

Exam Tip: Note that the ground state corresponds to \(n = 1\) and the second excited state corresponds to \(n = 3\) (not 2).

Q13. Sample A and B having the half lives 1h and 2h respectively. Initially both are having the equal number of undecayed nuclii. Find the ratio of their rate of reaction after 2hour. (3)
Answer: Let the initial number of undecayed nuclei in both samples be \(N_0\).
Decay constant \(\lambda\) is related to half-life \(T_{1/2}\) by:
\(\lambda = \frac{\ln 2}{T_{1/2}}\)
For sample A (\(T_{1/2A} = 1\) h):
\(\lambda_A = \frac{\ln 2}{1} = \ln 2\)
For sample B (\(T_{1/2B} = 2\) h):
\(\lambda_B = \frac{\ln 2}{2}\)
The rate of disintegration (activity \(R\)) at time \(t = 2\) hours is given by \(R = \lambda N\), where number of undecayed nuclei at time \(t\) is \(N = N_0 e^{-\lambda t}\).
For sample A after \(t = 2\) hours:
\(N_A = N_0 e^{-\lambda_A (2)} = N_0 e^{-2\ln 2} = N_0 \left(\frac{1}{2}\right)^2 = \frac{N_0}{4}\)
Rate of reaction for A:
\(R_A = \lambda_A N_A = (\ln 2) \left(\frac{N_0}{4}\right)\)
For sample B after \(t = 2\) hours:
\(N_B = N_0 e^{-\lambda_B (2)} = N_0 e^{-\left(\frac{\ln 2}{2}\right)(2)} = N_0 e^{-\ln 2} = \frac{N_0}{2}\)
Rate of reaction for B:
\(R_B = \lambda_B N_B = \left(\frac{\ln 2}{2}\right) \left(\frac{N_0}{2}\right) = (\ln 2) \left(\frac{N_0}{4}\right)\)
Ratio of their rates of reaction:
\(\frac{R_A}{R_B} = \frac{(\ln 2) \left(\frac{N_0}{4}\right)}{(\ln 2) \left(\frac{N_0}{4}\right)} = \frac{1}{1}\)
Therefore, the ratio is \(1:1\).
In simple words: After calculating the remaining nuclei and decay constants for both samples at 2 hours, their activities turn out to be equal.

Exam Tip: Activity \(R = \lambda N\) represents the rate of radioactive decay or reaction rate.

Q14. A heavy nucleus X of mass number 240 and binding energy per nucleon 7.6MeV is splitted in to two fragments Y and Z of mass number 110 and 130. Binding energy per nucleon of Y and Z is 8.5MeV per nucleon. Calculate the energy released per fission. (3)
Answer: Total binding energy of parent nucleus X:
\(\text{BE}_X = 240 \times 7.6 \text{ MeV} = 1824 \text{ MeV}\)
Total binding energy of daughter fragments Y and Z:
\(\text{BE}_Y = 110 \times 8.5 \text{ MeV} = 935 \text{ MeV}\)
\(\text{BE}_Z = 130 \times 8.5 \text{ MeV} = 1105 \text{ MeV}\)
Total binding energy of products = \(935 + 1105 = 2040 \text{ MeV}\)
Energy released per fission (\(Q\)) is the difference between total binding energy of products and reactants:
\(Q = \text{Total BE of products} - \text{Total BE of reactant}\)
\(Q = 2040 \text{ MeV} - 1824 \text{ MeV} = 216 \text{ MeV}\)
In simple words: Energy released equals the total binding energy of the resulting fragments minus the binding energy of the original nucleus.

Exam Tip: Always multiply the binding energy per nucleon by the mass number to find the total binding energy before subtracting.

Q15. Stopping potential vs frequency graph for two metals has been given.
(A) When the same radiation falls on both the metal surfaces the stopping potential of emitted electrons are found to be \(V_1\) and \(V_2\). Which one is greater?
(B) Find the slope in terms of \(V_1, V_2, \nu_{01}, \nu_{02}\) (3)
Answer: (A) From Einstein's photoelectric equation:
\(eV_0 = h\nu - \phi_0 \implies V_0 = \frac{h}{e}\nu - \frac{\phi_0}{e}\)
From the given graph, metal 1 has a smaller threshold frequency (\(\nu_{01}\) ) than metal 2 (\(\nu_{02}\)), meaning metal 1 has a smaller work function. For the same incident frequency \(\nu\), the metal with the smaller work function yields a higher stopping potential.
Therefore, \(V_1 > V_2\).
(B) The equation for the straight lines on the stopping potential vs frequency graph is:
\(V_0 = \frac{h}{e}\nu - \nu_0 \left(\frac{h}{e}\right)\)
The slope of the lines is universal and equal to \(\frac{h}{e}\).
Using points from the two lines at a common frequency \(\nu\) or using geometry of parallel lines:
\(\text{Slope} = \frac{V_1 - V_2}{\nu_{02} - \nu_{01}}\)
In simple words: The metal with the lower threshold frequency gives a higher stopping potential, and the slope can be expressed as the ratio of potential difference to threshold frequency difference.

Exam Tip: The slope of the stopping potential versus frequency graph for any metal is always equal to Planck's constant divided by electronic charge (\(h/e\)).