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NEET PHYSICSATOMSMedium

Question

In hydrogen spectrum, the shortest wavelength in the Balmer series is \lambda . The shortest wavelength in the Brackett series is:

A

16\lambda

B

2\lambda

C

4\lambda

D

9\lambda

Step-by-Step Solution

The shortest wavelength (series limit) corresponds to the transition from n2=∞n_2 = \infty. For the Balmer series, the transition is to n1=2n_1 = 2 . Using the Rydberg formula 1λ=R(1n12−1n22)\frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), we get 1λ=R(122−0)=R4\frac{1}{\lambda} = R \left(\frac{1}{2^2} - 0\right) = \frac{R}{4}, so λ=4R\lambda = \frac{4}{R}. For the Brackett series, the transition is to n1=4n_1 = 4 . The shortest wavelength λ′\lambda' is given by 1λ′=R(142−0)=R16\frac{1}{\lambda'} = R \left(\frac{1}{4^2} - 0\right) = \frac{R}{16}, so λ′=16R\lambda' = \frac{16}{R}. Comparing the two, λ′=4(4R)=4λ\lambda' = 4 \left(\frac{4}{R}\right) = 4\lambda.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ATOMS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSATOMShydrogenspectrumshortestwavelengthbalmer

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