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Atomic Structure: JEE Main Chemistry Question with Solution

Q: Do I need the exact numerical value of the Rydberg constant ?

No, taking the ratio of the two transition equations cancels completely, saving calculation time and avoiding rounding errors.

If wavelength of the first line of the Paschen series of hydrogen atom is

720\mathrm{~nm}

, then the wavelength of the second line of this series is

\mathrm{nm}

. (Nearest integer)

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Step-by-step solutionView

Correct answer

The wavelength of the second line of the Paschen series is 492 nm.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

For hydrogen atom ( $Z = 1$ ), the first line of the Paschen series has wavelength $\lambda_1 = 720\text{ nm}$ .

02find

Determine the wavelength of the second line of the Paschen series, $\lambda_2$ , rounded to the nearest integer.

03visualise

In the Paschen series, transitions terminate at $n_1 = 3$ . The first line corresponds to $n = 4 \to 3$ , and the second line corresponds to $n = 5 \to 3$ .

04strategise

Use the Rydberg formula $\frac{1}{\lambda} = R_H Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ . By taking the ratio $\frac{1/\lambda_2}{1/\lambda_1} = \frac{\lambda_1}{\lambda_2}$ , the constant $R_H$ cancels out directly.

05execute

For the first line ( $4 \to 3$ ): $\frac{1}{\lambda_1} = R_H \left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R_H \left(\frac{1}{9} - \frac{1}{16}\right) = R_H \left(\frac{7}{144}\right)$ For the second line ( $5 \to 3$ ): $\frac{1}{\lambda_2} = R_H \left(\frac{1}{3^2} - \frac{1}{5^2}\right) = R_H \left(\frac{1}{9} - \frac{1}{25}\right) = R_H \left(\frac{16}{225}\right)$ Dividing the two equations: $\frac{\lambda_1}{\lambda_2} = \frac{16/225}{7/144} = \frac{16 \times 144}{7 \times 225} = \frac{2304}{1575} = \frac{256}{175}$ Therefore: $\lambda_2 = \lambda_1 \times \frac{175}{256} = 720 \times \frac{175}{256} = \frac{126000}{256} \approx 492.1875\text{ nm}$

✓verify

Rounding $492.1875$ to the nearest integer gives $492\text{ nm}$ . As transition energy increases ( $5 \to 3$ is a larger energy drop than $4 \to 3$ ), wavelength must decrease: $\lambda_2 < 720\text{ nm}$ , which matches.

Hints that build this answer step by step

What are the principal quantum numbers $(n_1, n_2)$ for the first and second lines of the Paschen series?
First line: $n_1 = 3 \to n_2 = 4$ ; Second line: $n_1 = 3 \to n_2 = 5$
Using the Rydberg formula $\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ , what is the ratio $\frac{\lambda_2}{\lambda_1}$ ?
$\frac{\lambda_2}{\lambda_1} = \frac{\frac{1}{3^2} - \frac{1}{4^2}}{\frac{1}{3^2} - \frac{1}{5^2}} = \frac{7/144}{16/225}$
Given $\lambda_1 = 720\text{ nm}$ , what is $\lambda_2 = 720 \times \frac{7 \times 225}{16 \times 144}$ rounded to the nearest integer?
$492\text{ nm}$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Do I need the exact numerical value of the Rydberg constant

R_H

No, taking the ratio of the two transition equations cancels $R_H$ completely, saving calculation time and avoiding rounding errors.

Answer

The wavelength of the second line of the Paschen series is 492 nm.

Why each option works or fails

Step-by-step solution

given: For hydrogen atom ( $Z = 1$ ), the first line of the Paschen series has wavelength $\lambda_1 = 720\text{ nm}$ .
find: Determine the wavelength of the second line of the Paschen series, $\lambda_2$ , rounded to the nearest integer.
visualise: In the Paschen series, transitions terminate at $n_1 = 3$ . The first line corresponds to $n = 4 \to 3$ , and the second line corresponds to $n = 5 \to 3$ .
strategise: Use the Rydberg formula $\frac{1}{\lambda} = R_H Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ . By taking the ratio $\frac{1/\lambda_2}{1/\lambda_1} = \frac{\lambda_1}{\lambda_2}$ , the constant $R_H$ cancels out directly.
execute: For the first line ( $4 \to 3$ ): $\frac{1}{\lambda_1} = R_H \left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R_H \left(\frac{1}{9} - \frac{1}{16}\right) = R_H \left(\frac{7}{144}\right)$ For the second line ( $5 \to 3$ ): $n = 5 \to 3$ Dividing the two equations: $\frac{\lambda_1}{\lambda_2} = \frac{16/225}{7/144} = \frac{16 \times 144}{7 \times 225} = \frac{2304}{1575} = \frac{256}{175}$ Therefore: $\lambda_2 = \lambda_1 \times \frac{175}{256} = 720 \times \frac{175}{256} = \frac{126000}{256} \approx 492.1875\text{ nm}$
verify: Rounding $492.1875$ to the nearest integer gives $492\text{ nm}$ . As transition energy increases ( $5 \to 3$ is a larger energy drop than $4 \to 3$ ), wavelength must decrease: $\lambda_2 < 720\text{ nm}$ , which matches.

Atomic Structure: JEE Main Chemistry Question with Solution

What feels right?

The wavelength of the second line of the Paschen series is 492 nm.

Why each option works or fails

Solution

01given

02find

03visualise

04strategise

05execute

✓verify

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