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 $6.25 eV$. 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.)
Passage Based Questions
Passage
The $\beta$-decay process, discovered around 1900 , is basically the decay of a neutron $(n)$. In the laboratory, a proton $(p)$ and an electron $\left(e^{-}\right)$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., $n \rightarrow p+e^{-}+\bar{v} _e$, around 1930, Pauli explained the observed electron energy spectrum.
Assuming the anti-neutrino $\left(\bar{v} _e\right)$ 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 $0.8 \times 10^{6} eV$. The kinetic energy carried by the proton is only the recoil energy.
(2012)
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Solution:
- $\therefore \quad$ 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
$$ \begin{aligned} & =n(m V)=n P=n \sqrt{2 m q \Delta V} \\ & =\frac{200}{6.25 \times 1.6 \times 10^{-19}} \times \sqrt{2\left(9 \times 10^{-31}\right) \times 1.6 \times 10^{-19} \times 500} \\ & =24 \end{aligned} $$
42-43. Maximum kinetic energy of anti-neutrino is nearly $\left(0.8 \times 10^{6}\right) eV$.
