Question

The work function $(\varphi$) of some metals is listed below.
$[\text{Use}hc=1240\text{eV nm}]$

Metal	Work function $\left(\varphi \right)$ $\left(\text{eV}\right)$
$\text{Li}$	$2.4$
$\text{Cu}$	$4.8$
$\text{Ag}$	$4.3$
$\text{W}$	$4.75$

The metal(s) that shows photoelectric emission when illuminated by light of wavelength $300\text{nm}$.

• A. $\text{W}$
• B. $\text{Li}$
• C. $\text{Ag}$
• D. $\text{Cu}$

Answer 1

The correct option is B $\text{Li}$

Photoelectric emission takes place when,

${\text{Energy}}_{\text{incident radiation}}>\text{Work function}\left(\varphi \right)$

Here, the energy of incident radiation $\text{(in eV)}$ is given by,

$E=\frac{hc}{\lambda }$

$\Rightarrow E=\frac{1240\text{eV nm}}{300\text{nm}}$

$\Rightarrow E=4.133\text{eV}$

As, only ${\varphi }_{Li}<4.133\text{eV}$,

$\text{Li}$ shows photoelectric emission.

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

Why this question? Key concept: Photoelectric emission takes place when, ${\text{Energy}}_{\text{incident radiation}}>\text{Work function}\left(\varphi \right)$