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?

Answer 1

We know that,

$K.E.=h\nu -{\varphi }_0$

where $\varphi$ can be written as $h{\nu }_0$

When $\nu =2{\nu }_0$

$K.E{.}_1=\frac{m{v}_1^2}{2}=2h{\nu }_0-h{\nu }_0$

$V_1=\sqrt{\frac{2h{\nu }_0}{m}}$

Similarly we can find that

$V_2=\sqrt{\frac{4h{\nu }_0}{m}}$

Hence,

$\frac{V_1}{V_2}=1/2$