Question

The photoelectric thresshold wavelength of silver is $3250\times {10}^{-10}m$. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536\times {10}^{-10}m$ is approximately
(Given $h=4.14\times {10}^{-15}eV$ and $c=3\times {10}^{8}m{s}^{-1}$)

• A. $0.3\times {10}^{6}m{s}^{-1}$
• B. $6\times {10}^{5}m{s}^{-1}$
• C. $6\times {10}^{6}m{s}^{-1}$
• D. $61\times {10}^{3}m{s}^{-1}$

Answer 1

The correct option is B $6\times {10}^{5}m{s}^{-1}$

Kinetic Energy, $\frac{1}{2}m{v}^2=hc[\frac{1}{\lambda }-\frac{1}{{\lambda }_0}]$
Also we can write in terms of frequency,
$\frac{1}{2}m{v}^2=hv-h{v}_o$

$hv=\frac{4.14\times {10}^{-15}\times 3\times {10}^3}{2536\times {10}^{-10}}=4.897eV$

$h{v}_0=\frac{4.14\times {10}^{-15}\times 3\times {10}^3}{3250\times {10}^{-10}}=3.32eV$

$\frac{1}{2}m{v}^2=4.897-3.32=1.077eV$

$v=\sqrt{\frac{2\times 1.077\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}}$

$v=0.615\times {10}^{6}m/s$