SATHEE: Modern Physics 7 Question 39

Modern Physics 7 Question 39

41. 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 $6.25 e V$ . 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, $m_{e} = 9 \times 10^{- 31} k g$ and $e V = 1.6 \times 10^{- 19} J$ ) (2018 Adv.)

Passage Based Questions

Passage

The $β$ -decay process, discovered around 1900 , is basically the decay of a neutron $(n)$ . In the laboratory, a proton $(p)$ and an electron $(e^{-})$ are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e., $n \to p + e^{-} + {\bar{v}}_{e}$ , around 1930, Pauli explained the observed electron energy spectrum.

Assuming the anti-neutrino $({\bar{v}}_{e})$ to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is $0.8 \times 10^{6} e V$ . The kinetic energy carried by the proton is only the recoil energy.

(2012)

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Solution:

$∴$ Power $= n h f$

(where, $n =$ number of photons incident per second)

Since, $K E = 0, h f =$ work-funcition $W$

$\begin{aligned} 200 & = n W = n [6.25 \times 1.6 \times 10^{- 19}] \\ \Rightarrow n & = \frac{200}{1.6 \times 10^{- 19} \times 6.25} \end{aligned}$

As photon is just above threshold frequency $K E_{max}$ is zero and they are accelerated by potential difference of $500 V$ .

$\begin{aligned} ∴ K E_{f} & = q Δ V \\ \frac{P^{2}}{2 m} & = q Δ V \Rightarrow P = \sqrt{2 m q Δ V} \end{aligned}$

Since, efficiency is $100$ , number of electrons emitted per second $=$ number of photons incident per second.

As, photon is completely absorbed, force exerted

$\begin{aligned} = n (m V) = n P = n \sqrt{2 m q Δ V} \\ = \frac{200}{6.25 \times 1.6 \times 10^{- 19}} \times \sqrt{2 (9 \times 10^{- 31}) \times 1.6 \times 10^{- 19} \times 500} \\ = 24 \end{aligned}$