The ratio of the longest wavelength to the shortest wavelength \((\frac{\lambda_L}{\lambda_S})\) in Paschen series of hydrogen spectrum is:

This question was previously asked in
AAI ATC Junior Executive 27 July 2022 Shift 1 Official Paper
View all AAI JE ATC Papers >
  1. \(\frac{14}{7}\)
  2. \(\frac{8}{7}\)
  3. \(\frac{12}{7}\)
  4. \(\frac{16}{7}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{16}{7}\)
Free
AAI ATC JE Physics Mock Test
7.1 K Users
15 Questions 15 Marks 15 Mins

Detailed Solution

Download Solution PDF

Concept:

Hydrogen spectrum:

  • The hydrogen spectrum is an important piece of evidence to show the quantized electronic structure of an atom.
  • The hydrogen atoms of the molecule dissociate as soon as an electric discharge is passed through a gaseous hydrogen molecule.
  • It results in the emission of electromagnetic radiation initiated by the energetically excited hydrogen atoms.
  • The hydrogen emission spectrum comprises radiation of discrete frequencies.

​​F1 Madhuri Defence 20.12.2022 D8

  • First three spectral series in the hydrogen spectrum: Lyman, Balmer, and Paschen series
  • The formula for the wavelength, \(\frac{1}{λ}=R(\frac{1}{p^2}-\frac{1}{n^2})\) where, p =  higher energy level, n = lowest energy level, λ = wavelength
  • Here R = 109677 is called the Rydberg constant for the hydrogen atom

Calculation:

The longest wavelength in the Paschen series was obtained when p = 3, n = 4

For the longest wavelength, 

\(\frac{1}{λ_l}=R(\frac{1}{p^2}-\frac{1}{n^2})\)

\(\frac{1}{λ_l}=R(\frac{1}{3^2}-\frac{1}{4^2})\)

\(\frac{1}{λ_l}=R(\frac{1}{9}-\frac{1}{16})\)

\(λ _l=\frac{144}{7R}\)

The shortest wavelength in the Paschen series was obtained when p = 3, n = ∞ 

For the shortest wavelength, 

\(\frac{1}{λ_s}=R(\frac{1}{p^2}-\frac{1}{n^2})\)

\(\frac{1}{λ_s}=R(\frac{1}{3^2}-\frac{1}{∞^2})\)

\(λ _s=\frac{9}{R}\)

Ratio, \(\frac{λ_l}{λ _s}=\frac{144/7R}{9/R}\)

\(\frac{λ_l}{λ _s}=\frac{16}{7}\)

Latest AAI JE ATC Updates

Last updated on Jun 19, 2025

-> The AAI ATC Exam 2025 will be conducted on July 14, 2025 for Junior Executive.. 

-> AAI JE ATC recruitment 2025 application form has been released at the official website. The last date to apply for AAI ATC recruitment 2025 is May 24, 2025. 

-> AAI JE ATC 2025 notification is released on April 4, 2025, along with the details of application dates, eligibility, and selection process.

-> A total number of 309 vacancies are announced for the AAI JE ATC 2025 recruitment.

-> This exam is going to be conducted for the post of Junior Executive (Air Traffic Control) in the Airports Authority of India (AAI).

-> The Selection of the candidates is based on the Computer-Based Test, Voice Test and Test for consumption of Psychoactive Substances.

-> The AAI JE ATC Salary 2025 will be in the pay scale of Rs 40,000-3%-1,40,000 (E-1).

-> Candidates can check the AAI JE ATC Previous Year Papers to check the difficulty level of the exam.

-> Applicants can also attend the AAI JE ATC Test Series which helps in the preparation.

Get Free Access Now
Hot Links: teen patti party teen patti joy official teen patti bliss teen patti game online teen patti gold download