Question

When monochromatic light of wavelength $\lambda$ is incident on a metallic surface, the stopping potential for photoelectric current is $3V_0$. When same surface is illuminated with light of wavelength $2\lambda$, the stopping potential is $V_0$.
The threshold wavelength for this surface when photoelectric effect takes place is

• A. $\lambda$
• B. $2\lambda$
• C. $3\lambda$
• D. $4\lambda$

Answer 1

The correct option is D $4\lambda$

Kinetic energy of photoelectrons, $K.E=\frac{hc}{\lambda }-\varphi$

where work function, $\varphi =\frac{hc}{{\lambda }_o}$

Also stopping potential $\left(V\right)$ makes the photoelectrons to come to rest i.e $K.E=eV$

$⟹$$eV=hc\frac{({\lambda }_o-\lambda )}{{\lambda }_o\lambda }$

For incident wavelength $=\lambda$, stopping potential $V=3V_o$

$\therefore$ $e\left(3V_o\right)=hc\frac{({\lambda }_o-\lambda )}{{\lambda }_o\lambda }$ ................(1)

For incident wavelength $=2\lambda$, stopping potential $V=V_o$

$\therefore$ $e{V}_o=hc\frac{({\lambda }_o-2\lambda )}{{\lambda }_o\left(2\lambda \right)}$ ................(2)

Dividing (2) from (1), we get

$3=\frac{2({\lambda }_o-\lambda )}{({\lambda }_o-2\lambda )}$$⟹{\lambda }_o=4\lambda$