Question

If the threshold wavelength (${\lambda }_0$) for the ejection of an 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

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

As we know,

$W={\text{hv}}_0=\frac{hc}{{\lambda }_0}$

=$\frac{6.6\times {10}^{-34}Js\times 3\times {10}^{8}m}{330\times {10}^{-9}m}$

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

Hence, the correct option is $C$