Question

The work function of metal is $1$ $eV$. Light of wavelength $3000\mathring{A}$ is incident on this metal surface. The velocity of emitted photo electrons will be

• A. $10m/s$
• B. $1\times {10}^{3}m/s$
• C. $1\times {10}^{4}m/s$
• D. $1\times {10}^{6}m/s$

Answer 1

The correct option is D $1\times {10}^{6}m/s$

Energy of Light $=\frac{12400ev.\mathring{A}}{3000\mathring{A}}$

$\therefore$ Energy of Light $=\frac{12.4}{3}ev$

$\therefore$ Energy of Light $=4.133ev$

$\therefore$ Energy of emitted photo electron $=4.133-1$

$=3.133ev$

$\therefore \frac{1}{2}{m}_e{v}_e^2=3.133\times 1.6\times {10}^{-19}J$

$\therefore v_e^2=\frac{6.266\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}$

$\therefore v_e^2=1.10\times {10}^{12}$

$\therefore v_e={10}^{6}m/s$

Ans :- Velocity of emitted photo electron is ${10}^{6}m/s$