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)

1.03 eV

Here, $\lambda =600nm,{\lambda }^{\prime }=400nm,K_{max}^{\prime }=2K_{max}$

According to Einstein photoelectric equation

$K_{max}=\frac{hc}{\lambda }-{\varphi }_0$

and $2K_{max}=\frac{hc}{{\lambda }^{\prime }}-{\varphi }_0$

Dividing by we get

$2=\frac{\frac{hc}{{\lambda }^{\prime }}-{\varphi }_0}{\frac{hc}{\lambda }-{\varphi }_0}$ or $\frac{2hc}{\lambda }-2{\varphi }_0=\frac{hc}{{\lambda }^{\prime }}-{\varphi }_0$

$\frac{2hc}{\lambda }-2{\varphi }_0=\frac{hc}{{\lambda }^{\prime }}-{\varphi }_0$

Or $hc(\frac{2}{\lambda }-\frac{1}{{\lambda }^{\prime }})={\varphi }_0$

$\therefore {\varphi }_0=1240eVnm(\frac{2}{600nm}-\frac{1}{400nm})=1.03eV$