Home Maths JEE - Advanced Previous Year Paper JEE-Advanced-2018-Paper-2 In a photoelectric experiment a parallel bea…
Maths JEE - Advanced Previous Year Paper JEE-Advanced-2018-Paper-2 Numeric Response
Published on: August 13, 2026

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 x 10 -4 N due to the impact of the electrons. The value of n is__________.Mass of the electron m e = 9 x 10 -31 kg and 1.0 eV = 1.6 x 10 -19 J.

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The correct answer is:
24

(24)

No. of photoelectrons emitted per second

= 2x10 20 photoelectrons

Momentum of each electron before striking the anode

= 1.2 x 10' 23 kg m/s

The force experienced by the anode is

F = NP = 2 x 10 20 x 1.2 x 10 -23 = 24 x 10 -4 N

n = 24.00

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