Question

When a piece of metal is illuminated by monochromatic light of wavelength $\lambda , \, $ then stopping potential is $3V_{s}.$ When the same surface is illuminated by the light of wavelength $2\lambda $ , then stopping potential becomes $V_{s}.$ The value of threshold wavelength for photoelectric emission will be

• A. $4\lambda $
• B. $8\lambda $
• C. $\frac{4}{3}\lambda $
• D. $6\lambda $

Answer 1

Answer: (A)

According to Einstein's photoelectric equation
$eV=hc\left[\frac{1}{\lambda } - \frac{1}{\lambda _{0}}\right]$
Ist case $3eV_{s}=hc\left[\frac{1}{\lambda } - \frac{1}{\lambda _{0}}\right] \, $ ...(i)
IInd case $eV_{s}=hc$ $\left[\right.\frac{1}{2 \lambda }-\frac{1}{\lambda _{0}}\left]\right.$ ...(ii)
Dividing Eq. (i)by Eq. (ii), we get
$\lambda _{0}=4\lambda $