JEE Physics Dual Nature Of Radiation and Matter MCQs Set 10

Answer: (a) only one

Explanation:
1. In the photoelectric effect one photon is absorbed by one electron, and its whole energy is transferred to that single electron.
2. Hence at most one electron is emitted per photon.

Teacher's Note:
1. The interaction is a one-photon, one-electron event.
2. The number of photoelectrons per second depends on the number of photons per second, not on one photon.

Answer: (b) 4

Explanation:
1. Stopping potential is defined by \( eV_0=K_{max} \).
2. Here \( K_{max}=4\ \text{eV} \), so \( V_0=4\ \text{V} \).

Teacher's Note:
1. The stopping potential in volts equals the maximum kinetic energy in electron-volts.
2. The work function (2 eV here) is not needed once K_max is given.

Answer: (b) 0.04%

Explanation:
1. Energy of one photon \( =\frac{hc}{\lambda}=\frac{1240}{5000\ \text{Å}}\text{ eV}=2.48\ \text{eV}=3.97\times10^{-19}\ \text{J} \).
2. Photons per second \( =\frac{10^{-3}}{3.97\times10^{-19}}=2.52\times10^{15} \).
3. Electrons per second \( =\frac{0.16\times10^{-6}}{1.6\times10^{-19}}=10^{12} \).
4. Percentage \( =\frac{10^{12}}{2.52\times10^{15}}\times100\approx0.04\% \).

Teacher's Note:
1. Efficiency of emission is electrons per second divided by photons per second.
2. Convert the current to electrons per second using I = ne.

Answer: (c) Harmonic progression

Explanation:
1. \( V=\frac{hc}{e}\cdot\frac1\lambda-\frac{\phi}{e} \), so V is a linear function of \( \frac1\lambda \).
2. If V1, V2, V3 are in AP, then \( \frac{1}{\lambda_1},\frac1{\lambda_2},\frac1{\lambda_3} \) are in AP.
3. Reciprocals in AP means the wavelengths are in HP.

Teacher's Note:
1. A linear relation preserves arithmetic progression.
2. Numbers whose reciprocals are in AP are said to be in harmonic progression.

Answer: (c) λe > λp

Explanation:
1. \( \lambda=\frac{h}{\sqrt{2mqV}} \).
2. Same charge magnitude and same V, so \( \lambda\propto\frac{1}{\sqrt m} \).
3. The electron is much lighter than the proton, so λe > λp.

Teacher's Note:
1. For particles with equal charge accelerated through the same V, wavelength depends inversely on the square root of the mass.
2. The lighter particle has the longer wavelength.

Answer: (d) \( 10^{12}h \)

Explanation:
1. \( p=\frac h\lambda=\frac{h}{0.01\times10^{-10}}=\frac{h}{10^{-12}}=10^{12}h \).

Teacher's Note:
1. Convert the wavelength to metres first: 1 Å = 10⁻¹⁰ m.
2. Momentum of a photon is h divided by its wavelength.

Answer: (b) 4λ

Explanation:
1. Let \( E=\frac{hc}{\lambda} \). Then \( E=\phi+eV \) and \( \frac E2=\phi+\frac{eV}{3} \).
2. Subtracting: \( \frac E2=\frac{2eV}{3}\Rightarrow eV=\frac{3E}{4} \), so \( \phi=\frac E4 \).
3. Threshold wavelength \( \lambda_0=\frac{hc}{\phi}=4\frac{hc}{E}=4\lambda \).

Teacher's Note:
1. Write the photoelectric equation twice and eliminate eV to get the work function in terms of the photon energy.
2. The threshold wavelength is hc divided by the work function.

Answer: (b) 2 eV, 6200 Å

Explanation:
1. Photon energy \( =\frac{12400}{1240}=10\ \text{eV} \).
2. \( \phi=10-8=2\ \text{eV} \).
3. \( \lambda_0=\frac{12400}{2}=6200\ \text{Å} \).

Teacher's Note:
1. Use hc = 12400 eV Å for quick conversions between photon energy and wavelength.
2. Threshold wavelength corresponds to a photon whose energy equals the work function.

Answer: (a) 1.5 V

Explanation:
1. Photon energy at 200 nm \( =\frac{1240}{200}=6.2\ \text{eV} \).
2. \( eV_0=6.2-4.7=1.5\ \text{eV} \), so \( V_0=1.5\ \text{V} \).
3. (Check with the first case: 12.4 - 4.7 = 7.7 eV, which agrees.)

Teacher's Note:
1. Only the work function is needed; the first case simply confirms the data.
2. Use hc = 1240 eV nm.

Answer: (c) \( \frac{hc\lambda}{E\lambda+hc} \)

Explanation:
1. \( \frac{hc}{\lambda}=\phi+E \) and \( \frac{hc}{\lambda'}=\phi+2E \).
2. Subtract: \( \frac{hc}{\lambda'}=\frac{hc}{\lambda}+E=\frac{hc+E\lambda}{\lambda} \).
3. \( \lambda'=\frac{hc\lambda}{E\lambda+hc} \).

Teacher's Note:
1. Subtracting the two Einstein equations removes the unknown work function.
2. The new wavelength is shorter than λ, as expected for a more energetic photon.

Answer: (d) \( \frac{\lambda_0}{5000} \)

Explanation:
1. Mid visible wavelength \( \lambda_0\approx5500\ \text{Å} \).
2. For a 100 V electron, \( \lambda=\frac{12.27}{\sqrt{100}}=1.227\ \text{Å} \).
3. \( \frac{\lambda_0}{\lambda}=\frac{5500}{1.227}\approx4500 \), closest to 5000.

Teacher's Note:
1. Compare orders of magnitude: visible light is of the order of 5000 Å, 100 V electrons about 1 Å.
2. The ratio is therefore about 5000.

Answer: (c) λ > λ' > λ/2

Explanation:
1. \( \frac{hc}{\lambda}=\phi+K \), \( \frac{hc}{\lambda'}=\phi+2K=2\frac{hc}{\lambda}-\phi \).
2. So \( \frac{1}{\lambda'}=\frac2\lambda-\frac{\phi}{hc}<\frac2\lambda \), giving \( \lambda'>\frac\lambda2 \).
3. Also the energy rises, so λ' < λ. Hence λ > λ' > λ/2.

Teacher's Note:
1. Halving the wavelength would more than double the photon energy minus φ, so the new wavelength must be above λ/2.
2. It must still be below λ because the photon needs more energy.

Answer: (d) greater than 4K/3

Explanation:
1. \( \frac{hc}{\lambda}=\phi+K \). For wavelength 3λ/4 the photon energy is \( \frac43\frac{hc}{\lambda}=\frac43(\phi+K) \).
2. \( K'=\frac43(\phi+K)-\phi=\frac43K+\frac\phi3 \), which is greater than 4K/3.

Teacher's Note:
1. Photon energy scales as 1/λ, but the work function does not change.
2. So the extra kinetic energy exceeds the simple proportional increase.

Answer: (c) increase the maximum kinetic energy of the photoelectrons by a factor of more than two and have no effect on the magnitude of the photoelectric current

Explanation:
1. Maximum kinetic energy: \( K=hf-\phi \). With 2f, \( K'=2hf-\phi=2K+\phi>2K \).
2. Intensity (energy per unit area per second) is \( I=n\,hf \), where n is the number of photons per unit area per second. If I and f both double, n is unchanged.
3. The photocurrent depends on n, so it stays the same.

Teacher's Note:
1. Intensity measures energy per second, so doubling the photon energy while doubling the intensity leaves the photon rate unchanged.
2. Photocurrent is proportional to the number of photons arriving per second.

Answer: (c) K1 < K2/2

Explanation:
1. \( K_1=\frac{hc}{\lambda_1}-\phi=\frac{hc}{2\lambda_2}-\phi \) and \( K_2=\frac{hc}{\lambda_2}-\phi \).
2. So \( \frac{hc}{\lambda_2}=K_2+\phi \) and \( K_1=\frac{K_2+\phi}{2}-\phi=\frac{K_2-\phi}{2} \).
3. Since φ > 0, K1 < K2/2.

Teacher's Note:
1. Express hc/λ2 through K2 and φ, then halve it for λ1.
2. The work function subtracts a full amount, which reduces K1 below half of K2.

Answer: (c) 3.7 eV

Explanation:
1. Photon energy \( =\frac{1240}{200}=6.2\ \text{eV} \).
2. Maximum kinetic energy at emission \( =6.2-4.5=1.7\ \text{eV} \).
3. The collector is positive, so each electron gains 2.0 eV on its way: \( 1.7+2.0=3.7\ \text{eV} \).

Teacher's Note:
1. A positive collector accelerates the electrons, so it adds eV to their energy.
2. The maximum energy at emission is photon energy minus work function.

Answer: (b) v/2c

Explanation:
1. Equal wavelength means equal momentum p = h/λ.
2. Electron: \( E_e=\frac{p^2}{2m}=\frac{pv}{2} \). Photon: \( E_{ph}=pc \).
3. \( \frac{E_e}{E_{ph}}=\frac{pv/2}{pc}=\frac{v}{2c} \).

Teacher's Note:
1. Same wavelength gives same momentum for both particles.
2. Write each energy in terms of p and the speed.

Answer: (d) λ1 > λ2 or λ1 < λ2

Explanation:
1. A magnetic force is always perpendicular to the velocity, so the speed and wavelength of the second electron remain unchanged.
2. An electric force does work, so the speed of the first electron changes, but whether it speeds up or slows down depends on the direction of the field relative to the velocity.
3. Hence λ1 differs from λ2, and can be either larger or smaller.

Teacher's Note:
1. Magnetic fields do no work on a charge.
2. An electric field changes the speed, and the sign of the change depends on the geometry.

Answer: (d) 3 × 10⁷ W

Explanation:
1. For a perfectly absorbing surface the force is \( F=\frac{P}{c} \).
2. Equilibrium: \( \frac{P}{c}=mg \).
3. \( P=mgc=(0.01)(10)(3\times10^{8})=3\times10^{7}\ \text{W} \).

Teacher's Note:
1. Radiation force on an absorbing surface equals power divided by the speed of light.
2. Balance it against the weight.

Answer: (c) 3.33 × 10⁻⁵ N/m²

Explanation:
1. Intensity \( I=\frac{5\times10^{-3}}{10^{-6}}=5000\ \text{W/m}^2 \).
2. For complete reflection, \( P_{rad}=\frac{2I}{c}=\frac{2\times5000}{3\times10^{8}}=3.33\times10^{-5}\ \text{N/m}^2 \).

Teacher's Note:
1. A reflected photon reverses its momentum, so the momentum transferred is doubled.
2. Convert the area from mm² to m².

Answer: (b) E = 3.4 eV, λ = 6.6 × 10⁻¹⁰ m

Explanation:
1. In a Coulomb orbit, kinetic energy equals the magnitude of the total energy: E = 3.4 eV.
2. \( \lambda=\frac{12.27}{\sqrt{3.4}}\ \text{Å}=6.65\ \text{Å}=6.6\times10^{-10}\ \text{m} \).

Teacher's Note:
1. For bound orbits K = -E_total.
2. Use the electron formula λ (Å) = 12.27/√K (eV).

Answer: (d) 6 μN/m²

Explanation:
1. Intensity \( I=0.20\ \text{W/cm}^2=2000\ \text{W/m}^2 \).
2. Normal pressure for oblique incidence: \( p=\frac{I}{c}(1+\rho)\cos^2\theta \).
3. \( p=\frac{2000}{3\times10^{8}}\times1.8\times\frac12=6\times10^{-6}\ \text{N/m}^2=6\ \mu\text{N/m}^2 \).

Teacher's Note:
1. The absorbed part transfers momentum I cosθ/c per area; the reflected part gives an extra ρ times as much.
2. The normal component carries one more factor of cosθ because the beam is spread over a larger area.

Answer: (a) 0.18 μN

Explanation:
1. For a perfectly reflecting sphere, the net force equals that on a flat absorbing disc of the same cross-section: \( F=\frac{I}{c}\pi R^2 \).
2. \( F=\frac{7000}{3\times10^{8}}\times\pi(0.05)^2=2.33\times10^{-5}\times7.85\times10^{-3}\approx1.8\times10^{-7}\ \text{N} \), i.e. 0.18 μN.

Teacher's Note:
1. Reflection from different points of a sphere gives forces of different directions; after integrating, the net force equals I times the area of the shadow divided by c.
2. Convert I to W/m²: 0.70 W/cm² = 7000 W/m².

Answer: (d) none of these

Explanation:
1. By conservation of momentum the two fragments have momenta equal in magnitude: p1 = p2.
2. Since \( \lambda=\frac hp \), the wavelengths are equal and the ratio is 1, which is not among options (a) to (c).

Teacher's Note:
1. The initial momentum is zero, so the fragments move off with equal and opposite momenta.
2. De Broglie wavelength depends only on momentum.