Question

When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is $v_1$. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes $v_2$. If $v_2=x{v}_1$, the value of $x$ will be

Answer 1

Let us say the work function is $\varphi$
$\Rightarrow 2\varphi =\varphi +\frac{1}{2}m{v}_1^2\cdots \text{(1)}$
and $5\varphi =\varphi +\frac{1}{2}m{v}_2^2\cdots \text{(2)}$
From (1) and (2)
$\frac{v_2^2}{v_1^2}=\frac{4}{1}$ or $\frac{v_2}{v_1}=2$