Question

The work function of a metal is l eV. On making light of wavelength $3000A$ incident on this metals, the velocity of photoelectrons emitted from the metal, in m/s will be-

Answer 1

Energy of incident light$=\frac{hc}{\lambda }$

$=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{3000\times {10}^{-10}}=6.62\times {10}^{-19}$J

$hv=\varphi +\frac{1}{2}m{v}^2$

$\frac{hc}{\lambda }={\varphi }_o+\frac{1}{2}m{v}^2$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{hc}{\lambda }-{\varphi }_o$

$=6.62\times {10}^{-19}-1\times 1.6\times {10}^{-19}$

$=5.02\times {10}^{-19}$

$v^2=\frac{5.02\times {10}^{-19}\times 2}{9.1\times {10}^{-31}}=1.1\times {10}^{12}$

$v={10}^6$m/s

The velocity of the ejected photo electro is ${10}^6$ m/s.