If the electron in hydrogen atom jumps from third Bohr orbit to ground state directly and the difference between energies of the two states is radiated in the form of photons. If the work function of the material is $4 \cdot 1 \mathrm{eV}$, then stopping potential is nearly $\left[\text { Energy of electron in } \mathrm{n}^{\text {th }} \text { orbit }=\frac{-13 \cdot 6}{\mathrm{n}^2} \mathrm{eV}\right]$

The energy of an electron in the $n^{\text{th}}$ Bohr orbit is given by $E_n = \frac{-13.6}{n^2}$ eV.

For $n=3$, $E_3 = \frac{-13.6}{3^2} \approx -1.51$ eV, and for the ground state $n=1$, $E_1 = -13.6$ eV.

The energy difference when transitioning from $n=3$ to $n=1$ is $E_3 - E_1 = 12.09$ eV, which is the photon energy emitted.

In the photoelectric effect, $E_{\text{photon}} = W_0 + eV_s$, where $W_0 = 4.1$ eV.

Solving for the stopping potential, $eV_s = 12.09 - 4.1 = 7.99$ eV, giving $V_s = 7.99$ V.

This is approximately $8$ V, corresponding to option D.