Question

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

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

Answer 1

The correct option is D

$1\times {10}^{6}m/sec$

$E=W_0+K_{max};E=\frac{12375}{3000}=4.125eV$
$\Rightarrow K_{max}=E-W_0=4.125eV-1eV=3.125eV$
$\Rightarrow \frac{1}{2}m{v}_{max}^2=3.125\times 1.6\times {10}^{-19}J$
$\Rightarrow v_{max}=\sqrt{\frac{2\times 3.125\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}}=1\times {10}^{6}m/s$