Modern Physics Ques 73
- 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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Answer:
Correct Answer: 73.(a)
Solution:
Formula:
- According to the law of conservation of energy, i.e. Energy of a photon $(h \nu)=$ Work function $(\phi)+$ Kinetic energy of the photoelectron $(\frac{1}{2} m v _{\max }^{2})$
According to Einstein’s photoelectric effect theory
i.e.
$ \begin{alignedat} E = (KE)_{\text{max}} + \phi \\ \frac{h c}{\lambda} & =(KE) _{\max }+\phi \end{aligned} $
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 }+\phi $
$(KE) _{\max }$ for fastest emitted electron $=\frac{1}{2} m v^{2}-\phi$
$\Rightarrow \quad \frac{1}{2} m v^{\prime 2}=\frac{4}{3} (\frac{1}{2} m v^{2})+\frac{\phi}{3} $
$\text { i.e. } \quad v^{\prime}>v (\frac{4}{3})^{1/2}$
BETA
