Solve this following

Question:

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 $\frac{\mathrm{V}}{4}$. If the threshold

wavelength for the metallic surface is $n \lambda$ then

value of $\mathrm{n}$ will be

 

 

Solution:

$\frac{\mathrm{hc}}{\lambda}=\frac{\mathrm{hc}}{\lambda_{0}}+\mathrm{eV}$   ............(I)

$\frac{\mathrm{hc}}{3 \lambda}=\frac{\mathrm{hc}}{\lambda_{0}}+\frac{\mathrm{e} \cdot \mathrm{V}}{4}$ ..................(II)

(multiply by 4)

$\frac{4 \mathrm{hc}}{3 \lambda}=\frac{4 \mathrm{hc}}{\lambda_{0}}+\mathrm{eV}$ .................(III)

From (i) \& (iii)

$\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_{0}}=\frac{4 \mathrm{hc}}{3 \lambda}-\frac{4 \mathrm{hc}}{\lambda_{0}}$

$-\frac{\mathrm{hc}}{3 \lambda}=-\frac{3 \mathrm{hc}}{\lambda_{0}}$

$9 \lambda=\lambda_{0}$

$\mathrm{n}=9$

 

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