On a metal surface if light of wavelength λ falls, the stopping potential for emitted photo electrons is 3V₀. If light of wavelength 2λ falls, the stopping potential is V₀. Find the threshold wavelength

The correct option is 4λ.

Einstein's photoelectric equation relates the maximum kinetic energy of the emitted photoelectrons (which is equal to the product of electron charge e and the stopping potential V) to the energy of the incident light and the work function of the metal surface:

$eV=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$

where:
• h is Planck's constant
• c is the speed of light
• λ is the wavelength of the incident light
• λ₀ is the threshold wavelength of the metal surface
• V is the stopping potential

Let us write down the equations for the two cases given in the problem:

Case 1: When light of wavelength λ falls on the metal surface, the stopping potential is 3V₀.

$e\left(3V_0\right)=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$ — (Equation 1)

Case 2: When light of wavelength 2λ falls on the metal surface, the stopping potential is V₀.

$e{V}_0=\frac{hc}{2\lambda }-\frac{hc}{{\lambda }_0}$ — (Equation 2)

To eliminate eV₀ from the equations, we multiply Equation 2 by 3:

$3e{V}_0=\frac{3hc}{2\lambda }-\frac{3hc}{{\lambda }_0}$ — (Equation 3)

Now, equating the right-hand sides of Equation 1 and Equation 3:

$\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}=\frac{3hc}{2\lambda }-\frac{3hc}{{\lambda }_0}$

Dividing both sides by the common factor hc:

$\frac{1}{\lambda }-\frac{1}{{\lambda }_0}=\frac{3}{2\lambda }-\frac{3}{{\lambda }_0}$

Rearranging the terms to group the threshold wavelength λ₀ on one side and the incident wavelength λ on the other side:

$\frac{3}{{\lambda }_0}-\frac{1}{{\lambda }_0}=\frac{3}{2\lambda }-\frac{1}{\lambda }$

Simplifying both sides:

$\frac{2}{{\lambda }_0}=\frac{3}{2\lambda }-\frac{2}{2\lambda }$

$\frac{2}{{\lambda }_0}=\frac{1}{2\lambda }$

Cross-multiplying to solve for λ₀:

${\lambda }_0=4\lambda$

Therefore, the threshold wavelength is 4λ.