JEE Advanced · Physics · 26. Dual Nature
In a photoelectric experiment, a parallel beam of monochromatic light with the power of is incident on a perfectly absorbing cathode of work function . 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 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 due to the impact of the electrons. The value of is __________. Mass of the electron \(\text{m}_\text e =9 \times 10^{-31} kg\) and \(1.0 eV =1.6 \times\) \(10^{-19}\text{ J}.\)
- A 24
- B 25
- C 21
- D 27
Answer & Solution
Correct Answer
(A) 24
Step-by-step Solution
Detailed explanation
Power
number of photons per second
Since
As photon is just above threshold frequency is zero and they are accelerated by potential difference of
Since efficiency is 100%, number of electrons emitted is equal to number of photons falling per second as electrons are completely absorbed, force exerted
\(=\frac{200}{6.25 \times 1.6 \times 10^{-19}} \times\) \(\sqrt{2\left(9 \times 10^{-31}\right) \times 1.6 \times 10^{-19} \times 500} \)
\( =\frac{3 \times 200 \times 10^{-25} \times \sqrt{1600}}{6.25 \times 1.6 \times 10^{-19}}=\frac{2 \times 40}{6.25 \times 1.6} \times 10^{-4} \times 3=\) \(24 \times 10^{-4} \)
\( n=24\)
number of photons per second
Since
As photon is just above threshold frequency is zero and they are accelerated by potential difference of
Since efficiency is 100%, number of electrons emitted is equal to number of photons falling per second as electrons are completely absorbed, force exerted
\(=\frac{200}{6.25 \times 1.6 \times 10^{-19}} \times\) \(\sqrt{2\left(9 \times 10^{-31}\right) \times 1.6 \times 10^{-19} \times 500} \)
\( =\frac{3 \times 200 \times 10^{-25} \times \sqrt{1600}}{6.25 \times 1.6 \times 10^{-19}}=\frac{2 \times 40}{6.25 \times 1.6} \times 10^{-4} \times 3=\) \(24 \times 10^{-4} \)
\( n=24\)
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