Question 323: Dual Nature of Matter and Radiation

Question

The work function of metal ' $A$ ' and ' $B$ ' are in the ratio $1: 2$. If light of frequency ' $f$ ' and ' $2 f$ ' is incident on surface ' $A$ ' and ' $B$ ' respectively, then the ratio of kinetic energies of emitted photo electrons is

Accepted Answer

$$\begin{aligned}
& 	ext { For } \mathrm{A}, \mathrm{E}_{\mathrm{A}_{\max }}=\mathrm{h} v-\phi_{\mathrm{A}} \
& 	ext { For } \mathrm{B}, \mathrm{E}_{\mathrm{B}_{\max }}=\mathrm{h}(2 v)-\phi_{\mathrm{B}} \
& \frac{\mathrm{E}_{\mathrm{A}_{\max }}}{\mathrm{E}_{\mathrm{B}_{\max }}}= \frac{\mathrm{h} v-\phi_{\mathrm{A}}}{2 \mathrm{~h} v-\phi_{\mathrm{B}}} \
& 	ext { As } \frac{\phi_{\mathrm{A}}}{\phi_{\mathrm{B}}}=\frac{1}{2} \Rightarrow \phi_{\mathrm{B}}=2 \phi_{\mathrm{A}} \
& 	herefore \quad \frac{\mathrm{E}_{\mathrm{A}_{\max }}}{\mathrm{E}_{\mathrm{B}_{\max }}}=\frac{\frac{\mathrm{h} v-\phi_{\mathrm{A}}}{\phi_{\mathrm{A}}}}{\frac{2 \mathrm{~h} v-\phi_{\mathrm{B}}}{\phi_{\mathrm{A}}}}=\frac{\frac{\mathrm{h} v}{\phi_{\mathrm{A}}}-1}{\frac{2 \mathrm{~h} v}{\phi_{\mathrm{A}}}-2} \
&=\frac{\mathrm{h} v-\phi_{\mathrm{A}}}{\phi_{\mathrm{A}}} 	imes \frac{\phi_{\mathrm{A}}}{2\left(\mathrm{~h} v-\phi_{\mathrm{A}}ight)}=\frac{1}{2}
\end{aligned}$$

Answer

$1: 1$

Answer

$1: 2$

Answer

$1: 3$

Answer

$1: 4$

Dual Nature of Matter and Radiation

The work function of metal ' A ' and ' B ' are in the ratio 1: 2. If light of frequency ' f ' and ' 2 f ' is incident on surface ' A ' and ' B ' respectively, then the ratio of kinetic energies of emitted photo electrons is

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