Question

The ionisation energy of hydrogen atom in its ground state is $13.6\text{eV}$. The energy released (in $\text{eV}$) for the third line of Balmer series is:

• A. $-13.056$
• B. $-2.856$
• C. $-0.967$
• D. $-0.306$

Answer 1

The correct option is B $-2.856$

Given:
Ionisation energy of hydrogen atom in its ground state is $13.6\text{eV}$.

For the third line of Balmer series, $n_1=2$ and $n_2=5$

The energy released for the $3^{rd}$ line of Balmer series of hydrogen atom is:
$=-13.6\times (\frac{1}{n_1^2}-\frac{1}{n_2^2})=-13.6\times (\frac{1}{4}-\frac{1}{25})=-2.856\text{eV}$