Question: Q. 10. $X$-rays fall on a photosensitive surface to cause photoelectric emission. Assuming that the work function of the surface can be neglected, find the relation between the de-Broglie wavelength $(\lambda)$ of the electrons emitted to the energy $\left(E_{V}\right)$ of the incident photons. Draw the nature of the graph for $\lambda$ as a function of $E_{V}$. [Delhi Comptt. I, II, III 2014]
$$ \begin{array}{rlrl} \text {
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Solution:
Ans. } & E_{V} & =\phi_{0}+K_{\max } \ \text { As } & & \phi_{0} & =0 \ \Rightarrow & E_{\mathbf{V}} & =K_{\max } \ \Rightarrow & K_{\max } & =\frac{p^{2}}{2 m}=E_{\mathbf{V}} \ \Rightarrow & & p & =\sqrt{2 m E_{\mathrm{V}}} \end{array} $$
$\therefore$ Wavelength $(\lambda)$ of emitted electrons,
$$ \lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E_{\mathrm{V}}}} $$
[CBSE Marking Scheme, 2014]
