Question

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

• A. $2\times {10}^6\text{m}/\text{s}$
• B. $2\times {10}^7\text{m}/\text{s}$
• C. $8\times {10}^7\text{m}/\text{s}$
• D. $8\times {10}^6\text{m}/\text{s}$

Answer 1

The correct option is D $8\times {10}^6\text{m}/\text{s}$

Maximum kinetic energy is given by

$K_1=E_1-W=2h{v}_0-hv=h{v}_0$
$K_2=E_2-W=5h{v}_0-h{v}_0=4h{v}_0$

$K_2=4K_1$
$\Rightarrow {V_2}^2=4{V_1}^2$
$\Rightarrow V_2=2V_1$
$\therefore V_2=8\times {10}^{6}m/s$