Question

In hydrogen atom, energy of electron in ground state is \(13.6\,\text{eV}\), then energy of electron in second excited state is

• A. \(1.51\,\text{eV}\)
• B. \(3.4\,\text{eV}\)
• C. \(6.04\,\text{eV}\)
• D. \(13.5\,\text{eV}\)

Hint:

For hydrogen atom: \[ E_n = \frac{13.6}{n^2}\,\text{eV} \] Higher the principal quantum number \(n\), lower is the magnitude of energy.

Answer 1

The Correct Option is A

Step 1: Energy of electron in the \(n^{\text{th}}\) orbit of hydrogen atom is given by: \[ E_n = \frac{13.6}{n^2}\,\text{eV} \]
Step 2: Ground state corresponds to \(n = 1\). First excited state corresponds to \(n = 2\). Second excited state corresponds to \(n = 3\).
Step 3: Substitute \(n = 3\) in the formula: \[ E_3 = \frac{13.6}{3^2} = \frac{13.6}{9} = 1.51\,\text{eV} \]
Step 4: Hence, the energy of electron in the second excited state is: \[ \boxed{1.51\,\text{eV}} \]