Question

The work function for aluminium surface is 4.2 eV. The cut-off wavelength for the photoelectric effect is

• A. 2925 $\mathring{A}$
• B. 4200 $\mathring{A}$
• C. 2000 $\mathring{A}$
• D. 1000 $\mathring{A}$

Answer 1

Answer: (A)

If ${\lambda }_0$ is threshold wavelength, then work function
${\varphi }_0=\frac{hc}{{\lambda }_0}$
$\Rightarrow {\lambda }_0=\frac{hc}{{\varphi }_0}$
= $\frac{6.6\times 10^{-34}\times 3\times 10^8}{4.2\times 1.6\times 10^{-19}}$
$\approx 2955$ $\mathring{A}$