Question

If the ionisation energy for $H$ atom is $13.6eV$, then calculate the energy in eV when the atom is excited from the ground state to the first excited state?

• A. $3.6eV$
• B. $10.2eV$
• C. $13.75eV$
• D. $13.6eV$

Answer 1

The correct option is B $10.2eV$

Energy required to excite electron from ground state $(n=1)$ to first excited state $(n=2)$ will be given by,
$\Delta E=E_2-E_1\text{where}{E}_n=-13.6\times \frac{Z^2}{n^2}$
Hence,
$E_2=-13.6\times \frac{1^2}{2^2}$
$E_1=-13.6\times \frac{1^2}{1^2}$
Therefore,
$\Delta E=E_2-E_1=(-13.6\times \frac{1}{4})-(-13.6\times \frac{1}{1})$
$=13.6-3.4$ = $10.2eV$