Question 295: Dual Nature of Matter and Radiation

Question

When a certain metallic surface is illuminated with monochromatic light wavelength $\lambda$, the stopping potential for photoelectric current is $4 \mathrm{~V}_0$. When the same surface is illuminated with light of wavelength $3 \lambda$, the stopping potential is $\mathrm{V}_0$. The threshold wavelength for this surface for photoelectric effect is

Accepted Answer

Using Einstein's photoelectric equation,
$$\begin{aligned}
& \mathrm{h} v =\phi_0+\mathrm{KE}_{\max } \
	herefore \quad \frac{\mathrm{hc}}{\lambda} & =\phi_0+e\left(4 \mathrm{~V}_0ight)\quad	ext{.... (i)} \left(\because \mathrm{KE}_{\max }=\mathrm{eV}_{\mathrm{s}}ight)
\end{aligned}$$
Also, $\frac{\mathrm{hc}}{3 \lambda}=\phi_0+\mathrm{eV}_0\quad	ext{.... (ii)}$
Subtracting equation (i) from $4 	imes$ equation (ii) we get,
$$\left(\frac{4}{3}-1ight) \frac{\mathrm{hc}}{\lambda}=4 \phi_0-\phi_0 	ext { or } \phi_0=\frac{\mathrm{hc}}{9 \lambda}$$
But $\phi_0=\frac{\mathrm{hc}}{\lambda_0}$, where $\lambda_0$ is the threshold wavelength, hence $\lambda_0=9 \lambda$.
Hence, option (A) is correct.

Answer

$\quad 9 \lambda$

Answer

$\frac{\lambda}{9}$

Answer

$3 \lambda$

Answer

$\frac{\lambda}{3}$

Dual Nature of Matter and Radiation

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