Question

The work function of a metal is $4\text{eV}$. To emit a photoelectron of zero velocity from the surface of the metal, the wavelength of the incident light should be:

• A. $2500\stackrel{\circ }{A}$
• B. $3105\stackrel{\circ }{A}$
• C. $5600\stackrel{\circ }{A}$
• D. $1600\stackrel{\circ }{A}$

Answer 1

The correct option is B $3105\stackrel{\circ }{A}$

$W=4\text{eV}$
$h\nu =h{\nu }_0+\frac{1}{2}m{v}^2$
Given, $v=0$
So, $h\nu =h{\nu }_0$
$\Rightarrow \frac{hc}{\lambda }=\left(4\text{eV}\right)(1.6\times {10}^{-19})\text{J/eV}$
$\Rightarrow \lambda =\frac{(6.626\times {10}^{-34}\text{Js})(3\times {10}^8\text{m/s})}{(4\times 1.6\times {10}^{-19})\text{J}}$
$\Rightarrow \lambda =3105\times {10}^{-10}\text{m}$
$\Rightarrow \lambda =3105\stackrel{\circ }{A}$