Question

The light of two different frequencies whose photons have energies $3.8\text{eV}$ and $1.4\text{eV}$ respectively, illuminate a metallic surface whose work function is $0.6\text{eV}$ successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be

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

Answer 1

The correct option is C $2:1$

Energy of a photon is calculated by,

$E_{\text{photon}}=\varphi +\frac{1}{2}m{v}^2$

So, using the question's data,

$3.8=0.6+\frac{1}{2}m{v}_1^2$

$1.4=0.6+\frac{1}{2}m{v}_2^2$

$\Rightarrow \frac{v_1^2}{v_2^2}=\frac{3.2}{0.8}=\frac{4}{1}$

$\Rightarrow \frac{v_1}{v_2}=\frac{2}{1}$

Hence, $\left(B\right)$ is the correct answer.