JEE PYQ: Atomic Physics Question 24
Question 24 - 2020 (02 Sep Shift 1)
When radiation of wavelength $\lambda$ is used to illuminate a metallic surface, the stopping potential is $V$. When the same surface is illuminated with radiation of wavelength $3\lambda$, the stopping potential is $\frac{V}{4}$. If the threshold wavelength for the metallic surface is $n\lambda$, then value of $n$ will be ____.
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Answer: 9
Solution
$\frac{hc}{\lambda} = \phi + eV$ …(i). $\frac{hc}{3\lambda} = \phi + \frac{eV}{4}$ …(ii). From (i) $-$ (ii): $\frac{hc}{\lambda}\left(1 - \frac{1}{3}\right) = \frac{3}{4}eV$. So $\frac{hc}{\lambda} \cdot \frac{2}{3} = \frac{3}{4}eV \Rightarrow eV = \frac{8}{9}\frac{hc}{\lambda}$. $\frac{hc}{\lambda} = \frac{8}{9}\frac{hc}{\lambda} + \phi$. $\phi = \frac{hc}{9\lambda} = \frac{hc}{n\lambda}$, so $n = 9$.
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