Question

The photoelectric threshold wavelength for silver is $\lambda_{0}$ The energy of the electron ejected from the surface of silver by an incident wavelength $\lambda \left(\lambda <\lambda_{0}\right)$ will be

• A. $hc\left(\lambda_{0}-\lambda\right)$
• 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

Answer: (D)

$E =W+ K E$
$KE =E-W$
$=\frac{h c}{\lambda}-\frac{h c}{\lambda_{0}} $
$=h c\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right]= h c\left[\frac{\lambda_{0}-\lambda}{\lambda \lambda_{0}}\right]$