Question

The wave number of three spectral line of H-atom are given. The correct set of spectral lines belonging to Balmer series?

• A. $\frac{5R}{36}, \frac{3R}{16}, \frac{21R}{100}$
• B. $\frac{3R}{4}, \frac{3R}{16}, \frac{7R}{144}$
• C. $\frac{7R}{144}, \frac{3R}{16}, \frac{16R}{255}$
• D. $\frac{5R}{36}, \frac{3R}{16}, \frac{21R}{24}$

Hint:

Balmer series always has $n_1=2$. Simply plugging $n_2=3,4,5$ into the Rydberg formula generates the sequence.

Answer 1

The Correct Option is A

Balmer series involves transitions $n_2 \to n_1=2$.
Wave number $\bar{\nu} = R \left( \frac{1}{2^2} - \frac{1}{n_2^2} \right)$.
First line ($3 \to 2$): $\bar{\nu}_1 = R (\frac{1}{4} - \frac{1}{9}) = R \frac{5}{36}$.
Second line ($4 \to 2$): $\bar{\nu}_2 = R (\frac{1}{4} - \frac{1}{16}) = R \frac{3}{16}$.
Third line ($5 \to 2$): $\bar{\nu}_3 = R (\frac{1}{4} - \frac{1}{25}) = R \frac{21}{100}$.