Modern Physics Ques 180
- A photoelectric material having work-function $\phi _O$ is illuminated with light of wavelength $\lambda (\lambda<\frac{h c}{\phi _0})$. The fastest photoelectron has a de-Broglie wavelength $\lambda _d$. A change in wavelength of the incident light by $\Delta \lambda$ results in a change $\Delta \lambda _d$ in $\lambda _d$. Then, the ratio $\frac{\Delta \lambda _d}{\Delta \lambda}$ is proportional to
(a) $\frac{\lambda _d^{2}}{\lambda^{2}}$
(b) $\frac{\lambda _d}{\lambda}$
(c) $\frac{\lambda _d^{3}}{\lambda}$
(d) $\frac{\lambda _d^{3}}{\lambda^{2}}$
(2017 Adv.)
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Answer:
Correct Answer: 180.(d)
Solution:
Formula:
- According to photoelectric effect equation
$ \begin{array}{rlr} KE _{\max } & =\frac{h c}{\lambda}-\phi _0 & \\ \frac{p^{2}}{2 m} & =\frac{h c}{\lambda}-\phi _0 & {\left[KE=p^{2} / 2 m\right]} \\ \frac{\left(h / \lambda _d\right)^{2}}{2 m} & =\frac{h c}{\lambda}-\phi _0 & {[p=h / \lambda]} \end{array} $
Assuming small changes, differentiating both sides,
$ \frac{h^{2}}{2 m}(-\frac{2 d \lambda _d}{\lambda _d^{3}})=-\frac{h c}{\lambda^{2}} d \lambda , \quad \frac{d \lambda _d}{d \lambda} \propto \frac{\lambda _d^{3}}{\lambda^{2}} $
BETA
