JEE Physics Dual Nature Of Radiation and Matter MCQs Set 21

Answer: (c) 0.48 V

Explanation:
1. \( \lambda=5\times10^{-7} \) m \( =5000\ \text{\AA} \), so the photon energy is \( \dfrac{12375}{5000}=2.475 \) eV.
2. \( eV_0=2.475-2=0.475\approx0.48 \) eV, so \( V_0=0.48 \) V.

Teacher's Note:
1. Write Einstein's equation or the de Broglie relation first, and keep energies in eV with hc = 1240 eV nm.
2. Check that the answer is consistent with the threshold condition or a limiting case.

Answer: (a) \( 10^{11}\text{ and }5\text{ eV} \)

Explanation:
1. Power \( =16\times10^{-3}\times10^{-4}=1.6\times10^{-6} \) W. Photon energy \( =10\times1.6\times10^{-19}=1.6\times10^{-18} \) J, so photons per second \( =10^{12} \).
2. 10% produce photoelectrons: \( 10^{11} \) per second. \( K_{max}=10-5=5 \) eV.

Teacher's Note:
1. Write Einstein's equation or the de Broglie relation first, and keep energies in eV with hc = 1240 eV nm.
2. Check that the answer is consistent with the threshold condition or a limiting case.

Answer: (c) 1.8

Explanation:
1. A speed ratio of 2 means \( K_1=4K_2 \): \( 3.543-\varphi=4(2.296-\varphi) \).
2. \( 3\varphi=9.185-3.543=5.64\Rightarrow\varphi=1.88\approx1.8 \) eV.

Teacher's Note:
1. Write Einstein's equation or the de Broglie relation first, and keep energies in eV with hc = 1240 eV nm.
2. Check that the answer is consistent with the threshold condition or a limiting case.

Answer: (a) \( 6.4\times10^{-34}\ \text{J s} \)

Explanation:
1. \( eV_0=\dfrac{hc}\lambda-\varphi \) gives the slope \( \dfrac{h c}{e}=\dfrac{\Delta V_0}{\Delta(1/\lambda)}=\dfrac{2.0-1.0}{(1/0.3-1/0.4)\times10^6}=1.2\times10^{-6} \) V m.
2. \( h=\dfrac{1.2\times10^{-6}\times1.6\times10^{-19}}{3\times10^8}=6.4\times10^{-34} \) J s.

Teacher's Note:
1. Write Einstein's equation or the de Broglie relation first, and keep energies in eV with hc = 1240 eV nm.
2. Check that the answer is consistent with the threshold condition or a limiting case.

Answer: (c) 1.1 eV

Explanation:
1. Photon energy \( =13.6\left(\tfrac14-\tfrac19\right)=1.89 \) eV. Largest path means fastest electron: \( K_{max}=\dfrac{(eBr)^2}{2m}=\dfrac{(1.6\times10^{-19}\times3\times10^{-4}\times0.01)^2}{2\times9.1\times10^{-31}}=1.27\times10^{-19} \) J \( =0.79 \) eV.
2. \( \varphi=1.89-0.79=1.1 \) eV.

Teacher's Note:
1. Write Einstein's equation or the de Broglie relation first, and keep energies in eV with hc = 1240 eV nm.
2. Check that the answer is consistent with the threshold condition or a limiting case.

Answer: (a) 4

Explanation:
1. The stopping potential satisfies \( eV_0=K_{max} \).
2. \( V_0=\dfrac{4\text{ eV}}{e}=4 \) V (the work function is 2 eV).

Teacher's Note:
1. Write Einstein's equation or the de Broglie relation first, and keep energies in eV with hc = 1240 eV nm.
2. Check that the answer is consistent with the threshold condition or a limiting case.

Answer: (c) (ii) and (iv)

Explanation:
1. The stopping potential depends on frequency, not on intensity or distance, so it stays 0.6 V.
2. Intensity \( \propto\dfrac1{d^2} \): the distance is tripled, so the intensity and the saturation current fall to \( \dfrac19 \): \( \dfrac{18}9=2.0 \) mA.

Teacher's Note:
1. Write Einstein's equation or the de Broglie relation first, and keep energies in eV with hc = 1240 eV nm.
2. Check that the answer is consistent with the threshold condition or a limiting case.

Answer: (a) (iii) and (iv)

Explanation:
1. Emission needs \( \lambda\lt 5200\ \text{\AA} \), i.e. a frequency above the threshold. Infrared has longer wavelength, so it can never eject electrons whatever the power.
2. Ultraviolet has \( \lambda\lt 5200\ \text{\AA} \), so even the 1 W lamp ejects electrons (only the number changes with power).

Teacher's Note:
1. Write Einstein's equation or the de Broglie relation first, and keep energies in eV with hc = 1240 eV nm.
2. Check that the answer is consistent with the threshold condition or a limiting case.

Answer: (c) 7

Explanation:
1. \( K_{max}=\dfrac{1240}{200}-4.7=1.5 \) eV, so the sphere can reach \( V=1.5 \) V before emission stops.
2. \( Q=4\pi\varepsilon_0RV=\dfrac{0.01\times1.5}{9\times10^9}=1.67\times10^{-12} \) C, and \( N=\dfrac Qe=1.04\times10^7 \), so Z = 7.

Teacher's Note:
1. Write Einstein's equation or the de Broglie relation first, and keep energies in eV with hc = 1240 eV nm.
2. Check that the answer is consistent with the threshold condition or a limiting case.

Answer: (a) \( 2\times10^3\ \text{N/C} \)

Explanation:
1. Charge transferred \( =5\times10^7\times1.6\times10^{-19}=8\times10^{-12} \) C: plate A now has \( +8\times10^{-12} \) C and plate B has \( 33.7-8=25.7\times10^{-12} \) C... taking the net total as 33.7 pC shared as \( Q_A=+8 \) pC, \( Q_B=+25.7 \) pC.
2. Field between plates \( =\dfrac{Q_B-Q_A}{2A\varepsilon_0}=\dfrac{17.7\times10^{-12}}{2\times5\times10^{-4}\times8.85\times10^{-12}}=2\times10^3 \) N/C.

Teacher's Note:
1. Write Einstein's equation or the de Broglie relation first, and keep energies in eV with hc = 1240 eV nm.
2. Check that the answer is consistent with the threshold condition or a limiting case.

Answer: (c) \( 10^5 \)

Explanation:
1. Photons emitted per second \( =\dfrac{3.2\times10^{-3}}{5\times1.6\times10^{-19}}=4\times10^{15} \). The sphere intercepts the fraction \( \dfrac{\pi r^2}{4\pi d^2}=\dfrac{(8\times10^{-3})^2}{4\times0.64}=2.5\times10^{-5} \).
2. Incident photons \( =10^{11} \) per second, so photoelectrons \( =10^{11}/10^6=10^5 \) per second.

Teacher's Note:
1. Write Einstein's equation or the de Broglie relation first, and keep energies in eV with hc = 1240 eV nm.
2. Check that the answer is consistent with the threshold condition or a limiting case.

Answer: (a) 2.55 eV

Explanation:
1. \( E=K_{max}+\varphi=0.73+1.82 \).
2. \( E=2.55 \) eV.

Teacher's Note:
1. Write Einstein's equation or the de Broglie relation first, and keep energies in eV with hc = 1240 eV nm.
2. Check that the answer is consistent with the threshold condition or a limiting case.

Answer: (c) \( 0.814\ \text{m/s} \)

Explanation:
1. Photon momentum \( p=\dfrac Ec=\dfrac{2.55\times1.6\times10^{-19}}{3\times10^8}=1.36\times10^{-27} \) kg m/s.
2. \( v=\dfrac{p}{m}=\dfrac{1.36\times10^{-27}}{1.67\times10^{-27}}=0.814 \) m/s.

Teacher's Note:
1. Write Einstein's equation or the de Broglie relation first, and keep energies in eV with hc = 1240 eV nm.
2. Check that the answer is consistent with the threshold condition or a limiting case.

Answer: (a) The kinetic energy of the photoelectrons depends on the frequency of the radiation, not on its intensity

Explanation:
1. \( K_{max}=h\nu-\varphi \): it depends only on frequency and the metal.
2. Intensity changes only the number of photoelectrons (the photocurrent), not their energy.

Teacher's Note:
1. Write Einstein's equation or the de Broglie relation first, and keep energies in eV with hc = 1240 eV nm.
2. Check that the answer is consistent with the threshold condition or a limiting case.