Question 277: Dual Nature of Matter and Radiation

Question

When a photosensitive metal surface is illuminated with radiation of wavelength ' $\lambda_1$ ', the stopping potential is ' $\mathrm{V}_1$ '. If the same surface is illuminated with radiation of wavelength ' $3 \lambda_1$ ', the stopping potential is $\frac{\mathrm{V}_1}{6}$. The threshold wavelength for the photosensitive metal surface is

Accepted Answer

From $\frac{\mathrm{hc}}{\lambda}-\phi=\mathrm{eV}$,
We can write,
$$ \frac{\mathrm{hc}}{\lambda_1}=\mathrm{eV}_1+\phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$$
and
$$ \frac{\mathrm{hc}}{3 \lambda_1}=\frac{\mathrm{eV}_1}{6}+\phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$$
Equation (ii) can be rewritten as,
$$ \frac{2 \mathrm{hc}}{\lambda_1}=\mathrm{eV}_1+6 \phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$$
On subtracting equation (i) from (iii), we have $\frac{\mathrm{hc}}{\lambda_1}=5 \phi$ and $\phi=\frac{\mathrm{hc}}{\lambda_0}$
$$ 	herefore \quad \lambda_0=5 \lambda_1 $$

Answer

$\frac{3}{2} \lambda_1$

Answer

$2 \lambda_1$

Answer

$5 \lambda_1$

Answer

$6 \lambda_1$

Dual Nature of Matter and Radiation

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