What will be the first excitation energy of the electron in the hydrogen atom?

\(+9.84\text{ × }10^5\text{ J }mol^{-1}\)

The correct answer is option 2. \(+9.84\text{ × }10^5\text{ J }mol^{-1}\).

\(\text{The energy of the electron in the nth shell of the hydrogen atom is given by }\)

\(E_n\text{ = −}\frac{2\pi^2 me^4}{n^2h^2}\text{ = −}\frac{1.312 \text{ × }10^6}{n^2}\text{ J }mol^{-1}\)

\(\text{The first excitation energy is the amount of energy required to excite the electron from n=1 (ground state) to n=2 (first excited state), }\)

\(\Delta E \text{ = }E_2\text{ − }E_1\text{ = −}\frac{1.312 \text{ × }10^6}{2^2}\text{ − }(−\frac{1.312 \text{ × }10^6}{1^2})\)

\(\text{                                                          = −3.28 × }10^5\text{ + 13.12 × }10^5\text{ J }mol^{-1}\)

\(\text{                                                          =  +9.84 × }10^5\text{ J }mol^{-1}\)