Q3 / 150 · MHT CET 2026 (19 April Shift 1) · Chemistry
What is the wave number of lowest transition associated with Paschen series?
- A$\bar{v} = R_H\left(\dfrac{5}{36}\right)\ \text{cm}^{-1}$
- B$\bar{v} = R_H\left(\dfrac{36}{5}\right)\ \text{cm}^{-1}$
- C$\bar{v} = R_H\left(\dfrac{144}{7}\right)\ \text{cm}^{-1}$
- Correct answerD$R_H\left(\dfrac{7}{144}\right)\ \text{cm}^{-1}$
Solution
The wave number $\bar{v}$ for a transition in a hydrogen atom is given by the Rydberg formula:
$\bar{v} = R_H \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$
For the Paschen series, the lower energy level is $n_1 = 3$.
The lowest transition corresponds to the minimum energy difference, which occurs for the transition from the immediately higher energy level, $n_2 = 4$.
Substituting these values into the formula:
$\bar{v} = R_H \left( \dfrac{1}{3^2} - \dfrac{1}{4^2} \right)$
$\bar{v} = R_H \left( \dfrac{1}{9} - \dfrac{1}{16} \right)$
$\bar{v} = R_H \left( \dfrac{16 - 9}{144} \right)$
$\bar{v} = R_H \left( \dfrac{7}{144} \right) \text{cm}^{-1}$
Answer: $R_H\left(\dfrac{7}{144}\right)\ \text{cm}^{-1}$
$\bar{v} = R_H \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$
For the Paschen series, the lower energy level is $n_1 = 3$.
The lowest transition corresponds to the minimum energy difference, which occurs for the transition from the immediately higher energy level, $n_2 = 4$.
Substituting these values into the formula:
$\bar{v} = R_H \left( \dfrac{1}{3^2} - \dfrac{1}{4^2} \right)$
$\bar{v} = R_H \left( \dfrac{1}{9} - \dfrac{1}{16} \right)$
$\bar{v} = R_H \left( \dfrac{16 - 9}{144} \right)$
$\bar{v} = R_H \left( \dfrac{7}{144} \right) \text{cm}^{-1}$
Answer: $R_H\left(\dfrac{7}{144}\right)\ \text{cm}^{-1}$
Hence, the correct answer is option: 'D'.