Question

The work function of metal $A$ and $B$ are in the ratio $1:2$ . If light of frequencies $f$ and $2f$ are incident on the surfaces of $A$ and $B$ respectively, the ratio of the maximum kinetic energy of photo electrons emitted will be ( $f$ and $2f$ both frequency greater than threshold frequency of metal $A$ and $B$ )

• A. $1:1$
• B. $1:2$
• C. $1:3$
• D. $1:4$

Answer 1

Answer: (B)

Given $\frac{{\varphi }_A}{{\varphi }_B}=\frac{1}{2}$
$E_{K_1}=hf-{\varphi }_A$ $...(1)$
$E_{K_2}=h\left(2f\right)-{\left(\varphi \right)}_B$ $...(2)$
from Eq. $(1)$ and $(2)$
$E_{K_2}=2\left[E_{K_1}+{\varphi }_A\right]-2{\varphi }_A\left[\because {\varphi }_B=2{\varphi }_A\right]$
$\Rightarrow E_{K_2}=2E_{K_1}+2{\varphi }_A-2{\varphi }_A$
$\Rightarrow \frac{E_{K_1}}{E_{K_2}}=\frac{1}{2}$