Question

Work function for following metals given in the table.
Metal Work Function ${\varphi }_0\left(\text{eV}\right)$
$1$ $\text{K}$ $2.30$
$2$ $\text{Na}$ $2.75$
$3$ $\text{Mo}$ $4.17$
$4$ $\text{Ag}$ $4.7$
Their threshold wavelength in decreasing order must be

• A. ${\lambda }_K>{\lambda }_{Na}>{\lambda }_{Mo}>{\lambda }_{Ag}$
• B. ${\lambda }_K={\lambda }_{Na}={\lambda }_{Mo}={\lambda }_{Ag}$
• C. ${\lambda }_K<{\lambda }_{Na}<{\lambda }_{Mo}<{\lambda }_{Ag}$
• D. ${\lambda }_K>{\lambda }_{Na}={\lambda }_{Mo}<{\lambda }_{Ag}$

Answer 1

The correct option is A ${\lambda }_K>{\lambda }_{Na}>{\lambda }_{Mo}>{\lambda }_{Ag}$

Work function ${\varphi }_0=\frac{hc}{{\lambda }_0}$
${\lambda }_0=$Threshold wavelength is inversly proportional to work function
$\therefore$ Bigger the work function smaller is the threshold frequency.
Hence, option $\left(A\right)$ is the correct answer.

$$\begin{array}{c}\text{Why this Question ?}\\ \text{This question tests the basic concept;}\text{work function}\propto \frac{1}{\text{threshold wavelength}}\end{array}$$