Question 343: Dual Nature of Matter and Radiation

Question

From a metallic surface photoelectric emission is observed for frequencies $$v_1$$ and $$v_2\left(v_1 > v_2\right)$$ of the incident light. The maximum values of the kinetic energy of the photoelectrons emitted in the two cases are in the ratio $$1: \mathrm{x}$$. Hence the threshold frequency of the metallic surface is

Accepted Answer

Using Einstein's photoelectric equation,
$$\begin{array}{ll} 
& \mathrm{E}_{\mathrm{k}}=\mathrm{h} v-\phi_0 \
& \mathrm{E}_{\mathrm{k}}=\mathrm{h} v-\mathrm{hv}_0 \
	herefore & \mathrm{E}_{\mathrm{K}_1}=\mathrm{h}\left(v_1-v_0ight) 	ext { and } \mathrm{E}_{\mathrm{K}_2}=\mathrm{h}\left(v_2-v_0ight) \
& 	ext { Given } \frac{\mathrm{E}_{\mathrm{K}_1}}{\mathrm{E}_{\mathrm{K}_2}}=\frac{1}{\mathrm{x}} \
& \Rightarrow \frac{v_1-v_0}{v_2-v_0}=\frac{1}{\mathrm{x}} \
& \left(v_1-v_0ight) \mathrm{x}=v_2-v_0 \
& v_1 \mathrm{x}-v_0 \mathrm{x}=v_2-v_0 \
	herefore \quad & v_1 \mathrm{x}-v_2=v_0 \mathrm{x}-v_0 \
& v_1 \mathrm{x}-v_2=v_0(\mathrm{x}-1) \
	herefore \quad & v_0=\frac{v_1 \mathrm{x}-v_2}{\mathrm{x}-1}
\end{array}$$

Answer

$$\frac{v_1-v_2}{x}$$

Answer

$$\frac{v_1-v_2}{x-1}$$

Answer

$$\frac{x v_1-v_2}{x-1}$$

Answer

$$\frac{x v_2-v_1}{x-1}$$

Dual Nature of Matter and Radiation

Choose the correct answer: