The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes n=2 arrow n=3 and n=4 arrow n=6 transitions, respectively, is

Correct answer: D. (1)/(4) (1)/(λ)=RZ^2 ( 1 n_1^2 - 1 n_2^2 ) For 2 arrow 3 (1)/(λ_1)=R ×(1)^2((1)/(2^2)-(1)/(3^2)) (1) For 4 arrow 6 (1)/(λ_2)=R ×(1)^2((1)/(4^2)-(1)/(6^2)) (2) Divide (1) & (2) aligned & (λ_2)/(λ_1)=((1)/(2^2)-(1)/(3^2))/((1)/(4^2)-(1)/(6^2)) & (λ_2)/(λ_1)=(4)/(1) & (λ_1)/(λ_2)=(1)/(4) aligned

NEET 2025

Exam: NEET
Year: 2025
Subject: Chemistry
Chapter: Structure of Atom

NEET 2025 · Chemistry · Structure of Atom

The ratio of the wavelengths of the light absorbed by a Hydrogen...

The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes

\mathrm{n}=2 \rightarrow \mathrm{n}=3

and

\mathrm{n}=4 \rightarrow \mathrm{n}=6

transitions, respectively, is

Options

A $\frac{1}{36}$
B $\frac{1}{16}$
C $\frac{1}{9}$
D $\frac{1}{4}$

Correct answer

D. $\frac{1}{4}$

Step-by-step solution

\frac{1}{\lambda}=\mathrm{RZ}^2

\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)

For

2 \rightarrow 3 \Rightarrow \frac{1}{\lambda_1}=\mathrm{R} \times(1)^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \ldots(1)

For

4 \rightarrow 6 \Rightarrow \frac{1}{\lambda_2}=\mathrm{R} \times(1)^2\left(\frac{1}{4^2}-\frac{1}{6^2}\right) \ldots(2)

Divide (1) & (2)

\begin{aligned} & \frac{\lambda_2}{\lambda_1}=\frac{\frac{1}{2^2}-\frac{1}{3^2}}{\frac{1}{4^2}-\frac{1}{6^2}} \\ & \frac{\lambda_2}{\lambda_1}=\frac{4}{1} \\ & \therefore \frac{\lambda_1}{\lambda_2}=\frac{1}{4} \end{aligned}

Practice Structure of Atom on Quantrex Academy →

More from Structure of Atom