Question

What is the wave number of lowest transition associated with Paschen series?

• A. \(\overset{̄}{v} = R_H(\frac{5}{36}) \text{cm}^{-1}\)
• B. \(\overset{̄}{v} = R_H(\frac{36}{5}) \text{cm}^{-1}\)
• C. \(\overset{̄}{v} = R_H(\frac{144}{7}) \text{cm}^{-1}\)
• D. \(R_H(\frac{7}{144}) \text{cm}^{-1}\)

Hint:

Paschen series ends at n = 3, and its lowest-energy line is the transition from n = 4 to n = 3.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The Paschen series is the set of lines emitted when an electron falls to the level \(n_1 = 3\) from higher levels \(n_2 = 4, 5, 6,\dots\).
The "lowest transition" means the line with the smallest energy, hence the smallest wave number.

Step 2: Key Formula:
\[ \bar{v} = R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \]
The difference is smallest when \(n_2\) is closest to \(n_1\), so take \(n_2 = 4\).

Step 3: Calculation:
\[ \bar{v} = R_H\left(\frac{1}{3^2}-\frac{1}{4^2}\right) = R_H\left(\frac{1}{9}-\frac{1}{16}\right) = R_H\left(\frac{16-9}{144}\right) = R_H\,\frac{7}{144} \]

Step 4: Check the Other Options:
\(5/36\) equals \(\tfrac14-\tfrac19\), which is the first line of the Balmer series (3 to 2). \(36/5\) and \(144/7\) are reciprocals of these values, so they are not wave numbers at all. So (D) is correct.

Final Answer:
The lowest-energy Paschen line has wave number \(\tfrac{7}{144}R_H\ \text{cm}^{-1}\), option (D). \[ \boxed{\text{(D) } R_H\left(\tfrac{7}{144}\right)\text{cm}^{-1}} \]