Question

In a hydrogen atom, if the energy of an electron in ground state is $-13.6\text{eV}$, then its energy in the 2nd excited state is

• A. $-1.51\text{eV}$
• B. $3.4\text{eV}$
• C. $6.04\text{eV}$
• D. $-13.6\text{eV}$

Answer 1

The correct option is A $-1.51\text{eV}$

The $2^{nd}$ excited state will be the 3rd energy level.
For $2^{nd}$ excited state of hydrogen,
n = 3 and Z = 1
Energy of an electron in $n^{th}$ Bohr orbit is given by
$E_n=\frac{-13.6Z^2}{n^2}\text{eV}$
For 3rd orbit
$E_3=\frac{-13.6\times 1^2}{9}\text{eV}=-1.51\text{eV}$