In a photoelectric experiment, a parallel beam of monochromatic light with the power of 200    …

In a photoelectric experiment, a parallel beam of monochromatic light with the 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×10-4 N due to the impact of the electrons. The value of n is __________. Mass of the electron me=9×10-31 kg and 1.0 eV=1.6×10-19 J.

Solution

Power =Nhν
N= number of photons per second
Since KE=0,hν=ϕ
200=N6.25×1.6×10-19  Joule
N=2006.25×1.6×10-19
As photon is just above threshold frequency KEmax is zero and they are accelerated by potential difference of 500 V
KEf=q∆V
P22m=q∆V⇒P=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 nmv
= 2006.25×1.6×10-19×29×10-31×1.6×10-19×500
= 3×200×10-25×16006.25×1.6×10-19=2×406.25×1.6×10-4×3=24 × 10 −4
n=24 `

Asked in: JEE Advanced 2018 (Paper 2)

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