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JEE Advanced 2018Paper 2
Exam
JEE Advanced
Year
2018
Paper
Paper 2
Subject
Physics
Chapter
Dual Nature of Matter

JEE Advanced 2018 · Paper 2 · Physics · Dual Nature of Matter

In a photoelectric experiment, a parallel beam of monochromatic...

In a photoelectric experiment, a parallel beam of monochromatic light with the power of 200W200\text{W} is incident on a perfectly absorbing cathode of work function 6.25eV6.25eV. 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 500V500\text{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×10−4NF = n \times {10}^{- 4}N due to the impact of the electrons. The value of nn is __________. Mass of the electron me=9×10−31kgand1.0eV=1.6×10−19J{m}_{e} = 9 \times {10}^{- 31}kgand1.0eV = 1.6 \times {10}^{- 19}J.

Correct answer

0

Step-by-step solution

Power =Nhν= Nhν
N=N = number of photons per second
Since KE=0,hν=ϕKE = 0, hν = ϕ
200=N[6.25×1.6×10−19Joule]200 = N\left[6.25 \times 1.6 \times {10}^{- 19}Joule\right]
N=2006.25×1.6×10−19N = \dfrac{200}{6.25 \times 1.6 \times {10}^{- 19}}
As photon is just above threshold frequency K{E}_\max is zero and they are accelerated by potential difference of 500V500V
KEf=q∆VK{E}_{f} = q∆V
P22m=q∆V⇒P=2mq∆V\dfrac{{P}^{2}}{2m} = q∆V ⇒ P = \sqrt{2mq∆V}
Since efficiency is 100%, number of electrons emitted is equal to number of photons falling per second as electrons are completely absorbed, force exerted nmvnmv
=2006.25×1.6×10−19×2(9×10−31)×1.6×10−19×500= \dfrac{200}{6.25 \times 1.6 \times {10}^{- 19}} \times \sqrt{2\left(9 \times {10}^{- 31}\right) \times 1.6 \times {10}^{- 19} \times 500}
=3×200×10−25×16006.25×1.6×10−19=2×406.25×1.6×10−4×3=24×10−4= \dfrac{3 \times 200 \times {10}^{- 25} \times \sqrt{1600}}{6.25 \times 1.6 \times {10}^{- 19}} = \dfrac{2 \times 40}{6.25 \times 1.6} \times {10}^{- 4} \times 3 = 24\times {10}^{- 4}
n=24n = 24
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