Question

When a certain photosensitive surface is illuminated with monochromatic light of wavelength $\lambda$, the stopping potential for photoelectric current is $2V_0$. When the same surface is illuminated with monochromatic light of wavelength $2\lambda$, the stopping potential is $\frac{V_0}{2}$. The threshold wavelength for photoelectric emission is

• A. $1.5\lambda$
• B. $2.0\lambda$
• C. $2.5\lambda$
• D. $3\lambda$

Answer 1

The correct option is D $3\lambda$

We know that, $hv=h{v}_0+eV$
Since $v=\frac{c}{\lambda }$, we have $\frac{hc}{\lambda }=\frac{hc}{{\lambda }_0}+eV$, where ${\lambda }_0$ is the threshold wavelength.
For wavelength $\lambda$, $\frac{hc}{\lambda }=\frac{hc}{{\lambda }_0}+2e{V}_0$ .....(i)
And for wavelength $2\lambda$, $\frac{hc}{2\lambda }=\frac{hc}{{\lambda }_0}+\frac{e{V}_0}{2}$ .....(ii)
Eliminating $V_0$ from (i) and (ii), we get ${\lambda }_0=3\lambda$
Hence, the correct choice is (d).