Question

A light of wavelength $600 \,nm$ is incident on a metal surface. When light of wavelength $400 \,nm$ is incident, the maximum kinetic energy of the emitted photoelectrons is doubled. The work function of the metal is

• A. $1.03\,eV$
• B. $2.11\,eV$
• C. $4.14\,eV$
• D. $2.43\,eV$

Answer 1

Answer: (A)

Here, $\lambda=600\,nm$, $\lambda'=400\,nm$, $K'_{max}=2\,K_{max}$
According to Einsteins photoelectric equation
$K_{max}=\frac{hc}{\lambda}-\phi_{0} \ldots\left(i\right)$
and $2K_{max}=\frac{hc}{\lambda'}-\phi_{0} \ldots\left(ii\right)$
Dividing $\left(ii\right)$ by $\left(i\right)$ we get
$2=\frac{\left(\frac{hc}{\lambda'}\right)-\phi_{0}}{\left(\frac{hc}{\lambda}\right)-\phi_{0}} $ or $\frac{2hc}{\lambda}-2\phi_{0}=\frac{hc}{\lambda'}-\phi_{0}$
or $hc\left(\frac{2}{\lambda}-\frac{1}{\lambda'}\right)=\phi_{0}$ (Take hc $= 1240 \,eV nm)$
$\therefore \phi_{0}=1240\,eV\,nm \left(\frac{2}{600\,nm}-\frac{1}{400\,nm}\right)$
$=1.03\,eV$