Question

A photosensitive metallic surface has work function $h{\nu }_0.$ If photons of energy $2h{\nu }_0$ fall on this surface, the electrons come out with a maximum velocity of $4\times {10}^6{\text{ms}}^{-1}.$ When the photon energy is increased to $5h{\nu }_0$, then maximum velocity of photoelectrons will be

• A. $2\times {10}^7{\text{ms}}^{-1}$
• B. $8\times {10}^6{\text{ms}}^{-1}$
• C. $2\times {10}^6{\text{ms}}^{-1}$
• D. $8\times {10}^5{\text{ms}}^{-1}$

Answer 1

The correct option is B $8\times {10}^6{\text{ms}}^{-1}$

Given, ${\varphi }_0=h{\nu }_0$

From Einstein’s photoelectric equation,

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

$\frac{1}{2}m\times (4\times {10}^6{)}^2=2h{\nu }_0-h{\nu }_0.........(1)$

$\frac{1}{2}m\times v^2=5h{\nu }_0-h{\nu }_0............\left(2\right)$

Dividing (2) by (1),

$\frac{v^2}{(4\times {10}^6{)}^2}=\frac{4h{\nu }_0}{h{\nu }_0}$

$\Rightarrow v^2=4\times 16\times {10}^{12}$

$\Rightarrow v^2=64\times {10}^{12}$

$\therefore v=8\times {10}^6{\text{ms}}^{-1}$

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