Q. When a piece of metal is illuminated by a monochromatic light of wavelength $\lambda,\ $then stopping potential is $3V_{s}.$ When same surface is illuminated by 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) $6\lambda$
Explanation
(a) According to Einstein’s photoelectric equation
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ eV = hc\left\lbrack \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right\rbrack$
Ist case $3eV_{s} = hc\left\lbrack \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right\rbrack\ \ \ \ \ \ \ \ \ \ \ \ \ $…(i)
IInd case $eV_{s} = hc$ $\lbrack\frac{1}{2\lambda} - \frac{1}{\lambda_{0}}\rbrack$ …(ii)
Dividing Eq. (i)by Eq. (ii), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda_{0} = 4\lambda$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.