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 $\frac{\phi_{A}}{\phi_{B}}=\frac{1}{2}$
$E_{K_{1}}=hf-\phi_{A}$ $...\left(\right.1\left.\right)$
$E_{K_{2}}=h\left(2 f\right)-\left(\phi\right)_{B}$ $...\left(\right.2\left.\right)$
from Eq. $\left(\right.1\left.\right)$ and $\left(\right.2\left.\right)$
$E_{K_{2}}=2\left[E_{K_{1}} + \phi_{A}\right]-2\phi_{A }\left[\because \phi_{B} = 2 \phi_{A}\right]$
$\Rightarrow E_{K_{2}}=2E_{K_{1}}+2\phi_{A}-2\phi_{A}$
$\Rightarrow \frac{E_{K_{1}}}{E_{K_{2}}}=\frac{1}{2}$