Question

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

• A. $2\times {10}^{6}m/s$
• B. $2\times {10}^{7}m/s$
• C. $8\times {10}^{5}m/s$
• D. $8\times {10}^{6}m/s$

Answer 1

The correct option is D

$8\times {10}^{6}m/s$

$FromE=W_0+\frac{1}{2}m{v}_{max}^2$
$\Rightarrow 2h{v}_0=h{v}_0+\frac{1}{2}m{v}_1^2\Rightarrow h{v}_0=\frac{1}{2}m{v}_1^2$ . . . . .(i)
$and5h{v}_0=h{v}_0+\frac{1}{2}m{v}_2^2\Rightarrow 4h{v}_0=\frac{1}{2}m{v}_2^2$ . . . . . . (ii)
Dividing equation (ii) by (i) ${\left(\frac{v_2}{v_1}\right)}^2=\frac{4}{1}$
$\Rightarrow v_2=2v_1=2\times 4\times {10}^6=8\times {10}^{6}m/s$