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 1 eV , then stopping potential is nearly [ Energy of electron in n ^ th orbit = -13 6 n ^2 eV ]

Step-by-step solution

The energy of an electron in the n^ th Bohr orbit is given by E_n = -13.6 n^2 eV. For n=3 , E₃ = -13.6 3^2 -1.51 eV, and for the ground state n=1 , E₁ = -13.6 eV. The energy difference when transitioning from n=3 to n=1 is E₃ - E₁ = 12.09 eV, which is the photon energy emitted. In the photoelectric effect, E_ photon = W₀ + eV_s , where W₀ = 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 .