Modern Physics Ques 287
- 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 $F=n \times 10^{-4}$ $ N$ due to the impact of the electrons. The value of $n$ is (Take mass of the electron, $m _e=9 \times 10^{-31} kg$ and $eV=1.6 \times 10^{-19} J$ )
(2018 Adv.)
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Answer:
Correct Answer: 287.$(24)$
Solution:
Formula:
- $\therefore \quad$ Power $=n h f$
(where, $n=$ number of photons incident per second)
Since, $KE=0, h f=$ work-funcition $W$
$ \begin{aligned} 200 & =n W=n\left[6.25 \times 1.6 \times 10^{-19}\right] \\ \Rightarrow \quad n & =\frac{200}{1.6 \times 10^{-19} \times 6.25} \end{aligned} $
As photon is just above threshold frequency $K E _{\text {max }}$ is zero and they are accelerated by potential difference of $500$ $ V$.
$ \begin{aligned} \therefore \quad KE _f & =q \Delta V \\ \frac{P^{2}}{2 m} & =q \Delta V \Rightarrow P=\sqrt{2 m q \Delta V} \end{aligned} $
Since, efficiency is $100 \%$, number of electrons emitted per second $=$ number of photons incident per second.
As, photon is completely absorbed, force exerted
$ \begin{aligned} & =n(m V)=n P=n \sqrt{2 m q \Delta V} \\ & =\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} \\ & =24 \end{aligned} $
BETA
