Question

A hydrogen atom is excited from $n=1$ to $n=3$ state. The amount of energy absorbed by the atom will be

• A. $12.1eV$
• B. $25eV$
• C. $13.6eV$
• D. $-13.6eV$

Answer 1

Answer: (A)

$12.1eV$

Energy of $n^{th}$ excited state is
$E_n=-\frac{13.6}{n^2}$

Here, the hydrogen atom is excited from $n=1$ to $n=3$ state, hence the energy absorbed is equal to the difference between these two energy levels i.e.

$\Delta E=E_3-E_1$
$\Delta E=-\frac{13.6}{3^2}-(-\frac{13.6}{1^2})$

$\Delta E=-1.51+13.6$
$\Delta E=12.09\approx 12.1eV$