Question

If the threshold wavelength $\left({\lambda }_{\circ }\right)$ for the ejection of electrons from a metal is $330\text{nm}$, then the work function for the photoelectric emission is:

• A. $6\times {10}^{-10}\text{J}$
• B. $4\times {10}^{-18}\text{J}$
• C. $9\times {10}^{-19}\text{J}$
• D. $6\times {10}^{-19}\text{J}$

Answer 1

The correct option is D $6\times {10}^{-19}\text{J}$

Work function $\text{(W)}=h{\nu }_0$ = $\frac{hc}{{\lambda }_0}$
Where;
${\nu }_0=\text{Threshold frequency}$
${\lambda }_0=\text{Threshold wavelength}$
$\text{W}=\frac{(6.626\times {10}^{-34}\text{Js})(3\times {10}^8{\text{ms}}^{-1})}{(330\times {10}^{-9}\text{m})}$
$\text{W}=6\times {10}^{-19}\text{J}$