Question

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

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

Answer 1

Answer: (A)

Here, $\lambda =600nm$, ${\lambda }^{\prime }=400nm$, $K_{max}^{\prime }=2K_{max}$
According to Einsteins photoelectric equation
$K_{max}=\frac{hc}{\lambda }-{\varphi }_0\dots \left(i\right)$
and $2K_{max}=\frac{hc}{{\lambda }^{\prime }}-{\varphi }_0\dots \left(ii\right)$
Dividing $\left(ii\right)$ by $\left(i\right)$ we get
$2=\frac{\left(\frac{hc}{{\lambda }^{\prime }}\right)-{\varphi }_0}{\left(\frac{hc}{\lambda }\right)-{\varphi }_0}$ or $\frac{2hc}{\lambda }-2{\varphi }_0=\frac{hc}{{\lambda }^{\prime }}-{\varphi }_0$
or $hc\left(\frac{2}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)={\varphi }_0$ (Take hc $=1240eVnm)$
$\therefore {\varphi }_0=1240eVnm\left(\frac{2}{600nm}-\frac{1}{400nm}\right)$
$=1.03eV$