Question

The photoelectric threshold wavelength for Ag(silver) is ${\lambda }_0$. The energy of the electron ejected from the surface of silver by an incident wavelength $\lambda (\lambda <{\lambda }_0)$ will be:

• A. $hc({\lambda }_0-\lambda )$
  
• B. $\frac{hc}{{\lambda }_0-\lambda }$
  
• C. $\frac{h}{c}\left[\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right]$
  
• D. $hc\left[\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right]$

Answer 1

The correct option is D $hc\left[\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right]$

Equations for photoelectric effect
$W=W_0+K.E\frac{hc}{\lambda }=\frac{hc}{{\lambda }_0}+K.EK.E.=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}\Rightarrow [\frac{1}{\lambda }-\frac{1}{{\lambda }_0}]K.E=hc\left[\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right]$