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 $3e{V}_s=hc\left[\frac{1}{\lambda }-\frac{1}{{\lambda }_0}\right]$ ...(i)
IInd case $e{V}_s=hc$ $[\frac{1}{2\lambda }-\frac{1}{{\lambda }_0}]$ ...(ii)
Dividing Eq. (i)by Eq. (ii), we get
${\lambda }_0=4\lambda$