Find the wave number of the 4^ th line in the Lyman series of H spectrum. (R = Rydberg constant)

Correct answer: D. (24R)/25\ cm^ -1 For the Lyman series of the hydrogen spectrum, the transitions occur to the ground state, so n_1 = 1 . The 4^ th line corresponds to the transition from n_2 = 5 to n_1 = 1 . The wave number is given by the Rydberg formula: = R ( (1)/(n_1^2) - (1)/(n_2^2) ) Substituting n_1 = 1 and n_2 = 5 : = R ( (1)/(1^2) - (1)/(5^2) ) = R (1 - (1)/(25) ) = (24R)/(25) Answer: (24R)/25\ cm^ -1

MHT CET 202624 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 24 April
Shift: Morning Shift
Subject: Chemistry
Chapter: Structure of Atom

MHT CET 2026 · 24 April · Morning Shift · Chemistry · Structure of Atom

Find the wave number of the 4^ th line in the Lyman series of H...

Find the wave number of the

4^{\text{th}}

line in the Lyman series of H spectrum. (R = Rydberg constant)

Options

A $24/(25R)\ \text{cm}^{-1}$
B $25/(25R)\ \text{cm}^{-1}$
C $(25R)/24\ \text{cm}^{-1}$
D $(24R)/25\ \text{cm}^{-1}$

Correct answer

D. $(24R)/25\ \text{cm}^{-1}$

Step-by-step solution

For the Lyman series of the hydrogen spectrum, the transitions occur to the ground state, so

n_1 = 1

.

The

4^{\text{th}}

line corresponds to the transition from

n_2 = 5

n_1 = 1

.

The wave number

\bar{\nu}

is given by the Rydberg formula:

\bar{\nu} = R \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)<br><br>Substituting

n_1 = 1

and

n_2 = 5

:<br><br>\bar{\nu} = R \left( \dfrac{1}{1^2} - \dfrac{1}{5^2} \right)

\bar{\nu} = R \left(1 - \dfrac{1}{25} \right)<br><br>\bar{\nu} = \dfrac{24R}{25}<br><br>Answer:

(24R)/25\

\text{cm}^{-1}

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