Question

The threshold frequency for certain metal is ${\nu }_0.$ When light of frequency $2{\nu }_0$ is incident on it the maximum velocity of photo electrons is $4\times {10}^6{\text{ms}}^{-1}.$ If the frequency of incident radiation is increased to $4{\nu }_0$, then the maximum velocity of photo electrons will be-

• A. $2\times {10}^6{\text{ms}}^{-1}$
• B. $4\sqrt{3}\times {10}^6{\text{ms}}^{-1}$
• C. $6\times {10}^6{\text{ms}}^{-1}$
• D. $2\sqrt{3}\times {10}^6{\text{ms}}^{-1}$

Answer 1

The correct option is B $4\sqrt{3}\times {10}^6{\text{ms}}^{-1}$

From, Einstein's photo electric equation,

$h\nu =h{\nu }_0+\frac{1}{2}m{v}^2$

when, ${\nu }_1=2{\nu }_0$

$h{\nu }_1=h{\nu }_0+\frac{1}{2}m{v}_1^2$

$2h{\nu }_0-h{\nu }_0=\frac{1}{2}m{v}_1^2$

$h{\nu }_0=\frac{1}{2}m{v}_1^2......\left(1\right)$

When, ${\nu }_2=4{\nu }_0$

$h{\nu }_2=h{\nu }_0+\frac{1}{2}m{v}_2^2$

$4h{\nu }_0-h{\nu }_0=\frac{1}{2}m{v}_2^2$

$3h{\nu }_0=\frac{1}{2}m{v}_2^2.......\left(2\right)$

From (1) and (2) we get,

$\frac{3h{\nu }_0}{h{\nu }_0}=\frac{\frac{1}{2}m{v}_2^2}{\frac{1}{2}m{v}_1^2}$

$\frac{v_2^2}{v_1^2}=3$

$v_2=\sqrt{3}{v}_1$

$v_2=4\sqrt{3}\times {10}^6{\text{ms}}^{-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.