When photons of wavelength $$\lambda_1$$ are incident on an isolated sphere, the corresponding stopping potential is found to be $$V$$. When photons of wavelength $$\lambda_2$$ are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength $$\lambda_3$$ is used then find the stopping potential for this case:

For photo-emission we always begin with Einstein’s photoelectric equation, which relates the energy of an incident photon to the work function $$\phi$$ of the surface and to the maximum kinetic energy (expressed through the stopping potential) of the emitted electrons.

The equation is stated as

$$h\nu = \phi + eV_s,$$

where $$h$$ is Planck’s constant, $$\nu$$ the frequency of the light, $$V_s$$ the stopping potential and $$e$$ the electronic charge. Converting frequency into wavelength by using $$\nu = \dfrac{c}{\lambda}$$, the same relation becomes

$$\frac{hc}{\lambda} = \phi + eV_s.$$

We shall now write this equation for each of the three different wavelengths that appear in the question.

For the first wavelength $$\lambda_1$$ the stopping potential is given to be $$V$$, so

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

For the second wavelength $$\lambda_2$$ the stopping potential is thrice this value, i.e. $$3V$$. Therefore

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

We have two simultaneous equations in the two unknowns $$\phi$$ and $$eV$$. Subtracting equation (1) from equation (2) eliminates $$\phi$$:

$$\frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} = e(3V) - eV.$$

The right-hand side simplifies to $$e(2V)$$, hence

$$e(2V) = hc\!\left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right).$$

Dividing by 2 we obtain

$$eV = \frac{hc}{2}\!\left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right) \qquad (3).$$

Next we substitute the value of $$eV$$ from equation (3) back into equation (1) to determine the work function $$\phi$$:

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

We simplify the right-hand side term by term:

$$\phi = \frac{hc}{\lambda_1} - \frac{hc}{2\lambda_2} + \frac{hc}{2\lambda_1} = hc\!\left(\frac{1}{\lambda_1} + \frac{1}{2\lambda_1} - \frac{1}{2\lambda_2}\right) = \frac{hc}{2}\!\left(\frac{3}{\lambda_1} - \frac{1}{\lambda_2}\right).$$

Now we consider the third wavelength $$\lambda_3$$. Let the corresponding stopping potential be $$V_3$$. Applying the photoelectric equation once more gives

$$\frac{hc}{\lambda_3} = \phi + eV_3 \qquad (4).$$

We already have $$\phi$$, so we substitute it into equation (4):

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

Isolating $$eV_3$$ on the right-hand side we get

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

We now collect the terms inside the brackets carefully:

$$eV_3 = hc\!\left[\frac{1}{\lambda_3} - \frac{3}{2\lambda_1} + \frac{1}{2\lambda_2}\right].$$

Finally, dividing both sides by $$e$$ furnishes the required stopping potential for wavelength $$\lambda_3$$:

$$V_3 = \frac{hc}{e}\left[\frac{1}{\lambda_3} + \frac{1}{2\lambda_2} - \frac{3}{2\lambda_1}\right].$$

This expression coincides exactly with Option D.

Hence, the correct answer is Option D.