1. Use Einstein’s photoelectric equation

For an isolated sphere, the stopping potential $V_s$ satisfies

$eV_s = \frac{hc}{\lambda} - \phi$

where $\phi$ is the work function.

2. For wavelength $\lambda_1$

Given stopping potential is $V$:

$eV = \frac{hc}{\lambda_1} - \phi \qquad (1)$

3. For wavelength $\lambda_2$

Stopping potential is thrice, so it is $3V$:

$e(3V) = \frac{hc}{\lambda_2} - \phi \qquad (2)$

4. Eliminate $\phi$ or $V$

From (1):

$\phi = \frac{hc}{\lambda_1} - eV$

Substitute into (2):

$3eV = \frac{hc}{\lambda_2} - \left(\frac{hc}{\lambda_1} - eV\right)$

$3eV = \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} + eV$

$2eV = hc\left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right)$

Hence,

$eV = \frac{hc}{2}\left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right)$

5. For wavelength $\lambda_3$

Let the stopping potential be $V_3$. Then

$eV_3 = \frac{hc}{\lambda_3} - \phi$

Using $\phi = \frac{hc}{\lambda_1} - eV$ from (1),

$eV_3 = \frac{hc}{\lambda_3} - \left(\frac{hc}{\lambda_1} - eV\right)$

$eV_3 = hc\left(\frac{1}{\lambda_3} - \frac{1}{\lambda_1}\right) + eV$

Now substitute

$eV = \frac{hc}{2}\left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right)$

So,

$eV_3 = hc\left(\frac{1}{\lambda_3} - \frac{1}{\lambda_1}\right) + \frac{hc}{2}\left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right)$

$eV_3 = hc\left(\frac{1}{\lambda_3} + \frac{1}{2\lambda_2} - \frac{3}{2\lambda_1}\right)$

Therefore,

$V_3 = \frac{hc}{e}\left(\frac{1}{\lambda_3} + \frac{1}{2\lambda_2} - \frac{3}{2\lambda_1}\right)$

6. Match with options

This matches Option C.

$\boxed{\frac{hc}{e}\left(\frac{1}{\lambda_3} + \frac{1}{2\lambda_2} - \frac{3}{2\lambda_1}\right)}$