Question

If the threshold wavelength $\left({\lambda }^o\right)$ of 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 $6\times {10}^{-19}\text{J}$

Given information:
Threshold wavelength = $330\text{nm}$
Asked information: Work function

$\text{Work function}\left(w\right)=h{\nu }_o=\frac{hc}{{\lambda }_o}$
where, ${\nu }_o=\text{Threshold frequency}$
${\lambda }_o=\text{Threshold wavelength}$
$w=\frac{(6.626\times {10}^{-34}\text{J s})(3\times {10}^8{\text{m s}}^{-1})}{330\times {10}^{-9}\text{m}}$

$=6\times {10}^{-19}\text{J}$