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

The Balmer series of the hydrogen atom corresponds to electron transitions where the final energy level is \(n_f = 2\). The general formula for the wave number \(\bar{\nu}\) for transitions in hydrogen is given by:

\(\bar{\nu} = R \left( \frac{1}{n^2_f} - \frac{1}{n^2_i} \right)\)

Where \(R\) is the Rydberg constant, \(n_i\) is the initial energy level, and \(n_f\) is the final energy level.

For the Balmer series, \(n_f = 2\). Given the possible values of \(\bar{\nu}\):

1. \(\frac{5R}{36}\)
2. \(\frac{3R}{16}\)
3. \(\frac{21R}{100}\)
4. \(\frac{3R}{4}\)
5. \(\frac{7R}{144}\)
6. \(\frac{16R}{255}\)
7. \(\frac{21R}{24}\)

We need to check which values correspond to transitions to \(n_f = 2\).

Calculating the values for different \(n_i\) for \(n_f = 2\):

• For \(n_i = 3\), \(\bar{\nu} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R}{36}\)
• For \(n_i = 4\), \(\bar{\nu} = R \left( \frac{1}{4} - \frac{1}{16} \right) = \frac{3R}{16}\)
• For \(n_i = 5\), \(\bar{\nu} = R \left( \frac{1}{4} - \frac{1}{25} \right) = \frac{21R}{100}\)

The values \(\frac{5R}{36}\), \(\frac{3R}{16}\), and \(\frac{21R}{100}\) correspond to transitions in the Balmer series as calculated above.

Therefore, the correct set of wave numbers that belong to the Balmer series is:

\(\frac{5R}{36}, \frac{3R}{16}, \frac{21R}{100}\)