Question

When the light of frequency $2v_0$ (where $v_0$ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5v_0$, the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is

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

Answer 1

Answer: (B)

$E=W_0+\frac{1}{2}m{v}^2$
$h\left(2v_0\right)=h{v}_0+\frac{1}{2}m{v}_1^2$
$h{v}_0=\frac{1}{2}m{v}_1^2$ .....(i)
$h\left(5v_0\right)=h{v}_0+\frac{1}{2}m{v}_2^2$
$4h{v}_0=\frac{1}{2}m{v}_2^2$ .....(ii)
Divide (i) by (ii),
$\frac{1}{4}=\frac{v_1^2}{v_2^2}$
$\frac{v_1}{v_2}=\frac{1}{2}$