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JEE Advanced · Physics · 26. Dual Nature

In a photoelectric experiment, a parallel beam of monochromatic light with the power of 200W 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 500V 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 \(\text{m}_\text e =9 \times 10^{-31} kg\) and \(1.0 eV =1.6 \times\) \(10^{-19}\text{ J}.\)

  1. A 24
  2. B 25
  3. C 21
  4. D 27
Verified Solution

Answer & Solution

Correct Answer

(A) 24

Step-by-step Solution

Detailed explanation

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=qV
P22m=qVP=2mqV
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
\(=\frac{200}{6.25 \times 1.6 \times 10^{-19}} \times\) \(\sqrt{2\left(9 \times 10^{-31}\right) \times 1.6 \times 10^{-19} \times 500} \)
\( =\frac{3 \times 200 \times 10^{-25} \times \sqrt{1600}}{6.25 \times 1.6 \times 10^{-19}}=\frac{2 \times 40}{6.25 \times 1.6} \times 10^{-4} \times 3=\) \(24 \times 10^{-4} \)
\( n=24\)
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