Question

Find the threshold wavelength of light that would eject photoelectrons for a silver surface. The work function for silver is $4.8\text{eV}$. (Given $hc=1240\text{eV nm}$)

• A. $259\text{nm}$
• B. $220\text{nm}$
• C. $350\text{nm}$
• D. $540\text{nm}$

Answer 1

The correct option is A $259\text{nm}$

Work function, $\varphi =\frac{hc}{{\lambda }_0}$

Where, ${\lambda }_0\to \text{Threshold wavelength}$

$\Rightarrow {\lambda }_0=\frac{hc}{\varphi }$

$\Rightarrow {\lambda }_0=\frac{1240\text{eV nm}}{4.8\text{eV}}=258.33\text{nm}$

$\Rightarrow {\lambda }_0\approx 259\text{nm}$

Hence, option $\left(A\right)$ is correct.

Why this question? Key Note: Remember the difference between the threshold frequency and the threshold wavelength. The metal will emit electrons only if the frequency of the light incident on the metal is more than the threshold frequency. However, if the wavelength is more than the threshold wavelength, then the metal will not emit electrons.