Skip to content
dronacharya.ai
← JEE Advanced 2018 · Paper 2
PhysicsNumerical+3Question 13 of 54

JEE Advanced 2018 · Paper 2, Physics question 13

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%. 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×10−4F = n \times 10^{-4} N due to the impact of the electrons. The value of nn is __________. Mass of the electron me=9×10−31m_e = 9 \times 10^{-31} kg and 1.0 eV=1.6×10−191.0\ \mathrm{eV} = 1.6 \times 10^{-19} J.

Answer and worked solution

Official answer: 23.6

Electron rate. Each photon carries just ϕ=6.25\phi=6.25 eV and each releases one electron:
n=Phν=2006.25×1.6×10−19=2×1020 s−1.n=\frac P{h\nu}=\frac{200}{6.25\times1.6\times10^{-19}}=2\times10^{20}\ \mathrm{s^{-1}}.

Electron momentum at the anode, after accelerating from rest through 500 V:
p=2meV=2×9×10−31×1.6×10−19×500=1.44×10−46=1.2×10−23.p=\sqrt{2meV}=\sqrt{2\times9\times10^{-31}\times1.6\times10^{-19}\times500}=\sqrt{1.44\times10^{-46}}=1.2\times10^{-23}.

Force. The electrons are absorbed:
F=np=2×1020×1.2×10−23=2.4×10−3=24×10−4 N.F=np=2\times10^{20}\times1.2\times10^{-23}=2.4\times10^{-3}=24\times10^{-4}\text{ N}.

Answer: 24

Topics:Photoelectric effectImpulse and momentum
← Previous questionNext question →