Question

A metal sheet having a work function of $12.8\text{eV}$ is subjected to an electromagnetic radiation of wavelength $\left(\lambda \right)310\stackrel{\circ }{A}$. What will be the velocity of the photoelectrons having the maximum kinetic energy?

• A. $0$, no emission will occur
• B. $4.532\times {10}^6\text{m/s}$
• C. $3.09\times {10}^6\text{m/s}$
• D. $8.72\times {10}^6\text{m/s}$

Answer 1

The correct option is C $3.09\times {10}^6\text{m/s}$

$\lambda =310\stackrel{\circ }{A}$ and $W=12.8\text{eV}$
So, $h\nu =W+\frac{1}{2}m{v}^2$
$\frac{hc}{\lambda }=W+\frac{1}{2}m{v}^2$
$\Rightarrow \frac{(6.626\times {10}^{-34}\text{Js})(3\times {10}^8\text{m/s})}{(310\times {10}^{-10}\text{m})}=\left(12.8\text{eV}\right)(1.6\times {10}^{-19}\text{J/eV})+\frac{1}{2}m{v}^2$
$\Rightarrow 0.0641\times {10}^{-16}=(20.48\times {10}^{-19}\text{J})+\frac{1}{2}m{v}^2$
$\Rightarrow (64.1\times {10}^{-19}\text{J})-(20.48\times {10}^{-19}\text{J})=\frac{1}{2}m{v}^2$
$\Rightarrow \frac{(2\times 43.62\times {10}^{-19})\text{J}}{(9.1\times {10}^{-31}\text{kg})}=v^2[\because {\text{mass of e}}^{-}=9.1\times {10}^{-31}\text{kg}]$
$\Rightarrow v=3.09\times {10}^6\text{m/s}$