Question

Ionisation energy of hydrogen atom in ground state is 13.6 eV. The energy released (in eV) for third member of Balmer series is

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

Hint:

Balmer series always ends at \(n=2\); plug into \(13.6\left(\frac{1}{n_2^2} - \frac{1}{n_1^2}\right)\).

Answer 1

The Correct Option is B

Step 1: Identify the transition.
Third Balmer line corresponds to transition: \[ n = 5 \rightarrow n = 2 \] Step 2: Use hydrogen energy formula.
\[ E = 13.6\left(\frac{1}{2^2} - \frac{1}{5^2}\right) = 13.6\left(\frac{1}{4} - \frac{1}{25}\right) \] \[ = 13.6\left(\frac{25 - 4}{100}\right) = 13.6\left(\frac{21}{100}\right) = 2.856\ \text{eV} \]