Question

Using Bohr's formula of energy quantization , The ionisation potential of first excited state of hydrogen atom.

• A. 13.6 V
• B. 3.4 V
• C. 2.6 V
• D. 1.51 V

Answer 1

The correct option is B 3.4 V

According to Bohr’s theory

$E_n=\frac{-13.6}{n^2}eV$

At first excitation n=2

$E_2=\frac{-13.6}{2^2}=-3.4eV$

Potential energy at first excitation is $-3.4eV$