An electron of a hydrogen like atom, having $$Z = 4$$, jumps from $$4^{th}$$ energy state to $$2^{nd}$$ energy state. The energy released in this process, will be: (Given $$Rch = 13.6$$ eV, where $$R$$ = Rydberg constant, $$c$$ = Speed of light in vacuum, $$h$$ = Planck's constant)

We need to find the energy released when an electron in a hydrogen-like atom with $$Z = 4$$ jumps from the $$4^{th}$$ to the $$2^{nd}$$ energy state.

The energy of the $$n^{th}$$ level in a hydrogen-like atom is given by $$E_n = -\dfrac{Rch \cdot Z^2}{n^2}$$, where $$Rch = 13.6$$ eV. The energy released during a transition from level $$n_2$$ to level $$n_1$$ can be calculated by $$\Delta E = Rch \cdot Z^2 \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$$. Substituting $$n_1 = 2$$ and $$n_2 = 4$$ leads to $$\Delta E = 13.6 \times 4^2 \times \left(\dfrac{1}{4} - \dfrac{1}{16}\right) = 13.6 \times 16 \times \dfrac{3}{16} = 13.6 \times 3 = 40.8$$ eV.

The correct answer is Option D: 40.8 eV.