Question

If threshold wavelength $(\lambda^0)$ for ejection of electron from metal is $330\, nm$, then work function for the photoelectric emission is -

• A. $1.2 \times 10^{18}\, J$
• B. $1.2 \times 10^{-20}\, J$
• C. $6 \times 10^{-19}\, J$
• D. $6 \times 10^{-12}\, J$

Answer 1

Answer: (C)

$W=h \frac{C}{\lambda_{0}}$
$ \therefore W= \frac{6.6 \times 10^{-34}\times3\times10^{8}}{ 330\times10^{-9}}$
$= \frac{6.6 \times3\times10^{-18}}{33}$
$=0.6 \times10^{-18}$
$=6\times10^{-19} J $