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Physics · Atomic Physics

JEE Advanced 2018 — Paper 2 — Question 12

In a photoelectric experiment a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV . The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%100 \%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F=n×10−4 N\mathrm{F}=\mathrm{n} \times 10^{-4} \mathrm{~N} due to the impact of the electrons. The value of n is ____\_\_\_\_ .

Mass of the electron me=9×10−31 kg\mathrm{m}_{\mathrm{e}}=9 \times 10^{-31} \mathrm{~kg} and 1.0eV=1.6×10−19 J1.0 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}.

Answer: 24

Numerical answer — enter this value.

Step-by-step solution

No. of photoelectrons emitted per second

N=2006.25×1.6×10−19=2×1020\mathrm{N}=\frac{200}{6.25 \times 1.6 \times 10^{-19}}=2 \times 10^{20} photoelectron

Momentum of each electron before striking the anode

P=2×9×10−31×500×1.6×10−19=1.2×10−23 kg m/sP=\sqrt{2 \times 9 \times 10^{-31} \times 500 \times 1.6 \times 10^{-19}}=1.2 \times 10^{-23} \mathrm{~kg} \mathrm{~m} / \mathrm{s}

∴\therefore The force experienced by the anode is

F=NP=2×1020×1.2×10−23=24×10−4 N\mathrm{F}=\mathrm{NP}=2 \times 10^{20} \times 1.2 \times 10^{-23}=24 \times 10^{-4} \mathrm{~N}

∴n=24.00\therefore \mathrm{n}=24.00

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2018
Paper
Paper 2
Subject
Physics
Chapter
Atomic Physics
Topic
Photoelectric Effect

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