Question

The kinetic energy of the photoelectrons is $E$ when the incident wavelength is $\lambda /2$. The kinetic energy becomes $2E$ when the incident wavelength is $\lambda /3$. The work function of the metal is:

• A. $hc/\lambda$
• B. $2hc/\lambda$
• C. $3hc/\lambda$
• D. $hc/3\lambda$

Answer 1

The correct option is A $hc/\lambda$

Einstein equation for photoelectric effect : $KE=\frac{hc}{\lambda }-W$

where $KE=$ kinetic energy of the photoelectrons , $h=$Planck's constant , $c=$ velocity of light and $\lambda =$ wavelength incident light and $W=$ work function of metal.

When $KE=E,$ then $E=\frac{2hc}{\lambda }-W...\left(1\right)$

and when $KE=2E,$ then $2E=\frac{3hc}{\lambda }-W...\left(2\right)$

Using (1) , (2) becomes , $2(2hc/\lambda -W)=\frac{3hc}{\lambda }-W$

or $W=\frac{hc}{\lambda }$