Question

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

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

Answer 1

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

We know that,

$E=hv=\frac{h_c}{\lambda }=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{3\times {10}^{-3}}$

$E=6.6\times {10}^{-19}J$

we know that,

$E=\varphi +KE$

$6.6\times {10}^{-19}-1.6\times {10}^{-19}=\frac{1}{2}{mV}^2[1eV-16\times {10}^{-19}]$

$V^2=\frac{5\times {10}^{-19}\times 2.1.1\times {10}^{12}}{9.1\times {10}^{-31}}$

$V=1\times {10}^{6}m/s$