Question 342: Dual Nature of Matter and Radiation

Question

When a certain metal surface is illuminated with light of frequency $$v$$, the stopping potential for photoelectric current is $$\mathrm{V}_0$$. When the same surface is illuminated by light of frequency $$\frac{v}{2}$$, the stopping potential is $$\frac{\mathrm{V}_0}{4}$$, the threshold frequency of photoelectric emission is

Accepted Answer

$$\begin{aligned}
& \mathrm{eV}_0=\mathrm{h} v-\mathrm{h} v_0 \quad 	ext{..... (i)}\
& \frac{\mathrm{eV}_0}{4}=\frac{\mathrm{h} v}{2}-\mathrm{hv} v_0 \quad 	ext{..... (ii)}
\end{aligned}$$
Dividing equation (i) by equation (ii),
$$\begin{array}{ll} 
& 4=\frac{v-v_0}{\frac{v}{2}-v_0} \
	herefore \quad & 2 v-4 v_0=v-v_0 \
	herefore \quad & 3 v_0=v \
	herefore \quad & v_0=\frac{v}{3}
\end{array}$$

Answer

$$\frac{v}{6}$$

Answer

$$\frac{v}{3}$$

Answer

$$\frac{2 v}{3}$$

Answer

$$\frac{4 v}{3}$$

Dual Nature of Matter and Radiation

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