Question

The ionization potential of $H$ atoms is $13.6V$. The energy difference between $n=2$ and $n=3$ levels is nearest to

• A. $1.9eV$
• B. $2.3eV$
• C. $3.4eV$
• D. $4.5eV$

Answer 1

The correct option is A $1.9eV$

Energy is given by equation, $E_n=-\frac{13.6}{n^2}eV$

Therefore, $E_3-E_2=-\frac{13.6}{2^2}+\frac{13.6}{3^2}eV=1.9eV$