Modern Physics 2 Question 12
12. Radiation of wavelength $\lambda$, is incident on a photocell. The fastest emitted electron has speed $v$. If the wavelength is changed to $\frac{3 \lambda}{4}$, the speed of the fastest emitted electron will be
(2016 Main)
(a) $>v \frac{4}{3}^{1 / 2}$
(b) $<v \frac{4}{3}^{1 / 2}$
(c) $=v \frac{4}{3}^{1 / 2}$
(d) $=v \frac{3}{4}^{1 / 2}$
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Solution:
- According to the law of conservation of energy, i.e. Energy of a photon $(h \nu)=$ Work function $(\varphi)+$ Kinetic energy of the photoelectron $\frac{1}{2} m v _{\max }^{2}$
According to Einstein’s photoelectric emission of light i.e.
$$ \begin{aligned} E & =(KE) _{\text {max }}+\varphi \\ \frac{h c}{\lambda} & =(KE) _{\max }+\varphi \end{aligned} $$
As,
If the wavelength of radiation is changed to $\frac{3 \lambda}{4}$, then
$$ \Rightarrow \quad \frac{4}{3} \frac{h c}{\lambda}=\frac{4}{3}(KE) _{\max }+\frac{\varphi}{3}+\varphi $$
$(KE) _{\max }$ for fastest emitted electron $=\frac{1}{2} m v^{2}+\varphi$
$$ \begin{array}{ll} \Rightarrow & \frac{1}{2} m v^{\prime 2}=\frac{4}{3} \frac{1}{2} m v^{2}+\frac{\varphi}{3} \\ \text { i.e. } & v^{\prime}>v \frac{4}{3} \end{array} $$
