Question

A light source of wavelength $\lambda$ illuminates a metal and ejects photoelectrons with $(K.E{)}_{max}=1\text{eV}$.
Another light source of wavelength $\frac{\lambda }{3}$, ejects photoelectrons from same metal with $(K.E{)}_{max}=4\text{eV}$.
Find the value of work function of the metal.

• A. $1\text{eV}$
• B. $2\text{eV}$
• C. $0.5\text{eV}$
• D. None of these

Answer 1

The correct option is C $0.5\text{eV}$

$h\nu =h{\nu }_0+\frac{1}{2}m{v}^2$
$\frac{hc}{\lambda }=\omega +K.E$
Where,
$\omega =\text{work function}$
When the wavelength is $\lambda$,
$K.E=1\text{eV}$
$\therefore \frac{hc}{\lambda }=\omega +1...\left(1\right)$
When the wavelength is $\frac{\lambda }{3}$,
$K.E=4\text{eV}$
$\therefore \frac{3hc}{\lambda }=\omega +4...\left(2\right)$
On solving equations (1) and (2)
$\omega =0.5\text{eV}$