Question

A light source of wavelength $\lambda$ illuminates a metal and Ejects photo electron with $(KE{)}_{max}=1eV$. Another light source of wavelength $\frac{\lambda }{3}$,ejects photo-electrons from same metal with $(KE{)}_{max}=5eV$. Find the value of work function (eV) of metal.

Answer 1

$E=KE+W$

E is the total energy of the incident light source $(E=h\nu =\frac{hc}{\lambda })$

KE is the kinetic energy of the ejected photoelectron

W is the work function of the metal. It is constant for a particular metal.

• When light of wavelength $\lambda$ is incident on a metal then,
  

$\frac{hc}{\lambda }=1+W$

• When light of wavelength $\frac{\lambda }{3}$ is incident

$\frac{hc}{\lambda /3}=5+W$

$\frac{3hc}{\lambda }=5+W$

Putting the value of $\frac{hc}{\lambda }$, we get

$3(1+W)=5+W$

$2W=2$

$W=1eV$