
Balmer gives an equation for wavelength of visible radiation of $H$ spectrum as $\lambda = \dfrac{{k{n^2}}}{{{n^2} - 4}}$ .The value of $k$ in terms of Rydberg's constant $R$ is
(A) $R$
(B) $4R$
(C) $\dfrac{R}{4}$
(D) $\dfrac{4}{R}$
Answer
145.5k+ views
Hint: Balmer series occurs only when electrons shift from nth energy level to the second energy level. We substitute values of the energy level in Rydberg’s wavelength formula and find the relation between $k$ and Rydberg’s constant.
Formula used
Rydberg’s wavelength equation $\dfrac{1}{\lambda } = R(\dfrac{1}{{{n_f}}} - \dfrac{1}{{{n_i}}})$
Here,
Wavelength is represented by $\lambda $
Rydberg’s constant is represented by $R$
Initial and final energy levels are represented by ${n_i},{n_f}$ respectively
Complete step by step solution:
Rydberg’s equation for visible radiation of the hydrogen spectrum is given as
$\dfrac{1}{\lambda } = R(\dfrac{1}{{{n_f}^2}} - \dfrac{1}{{{n_i}^2}})$
For the Balmer series, the energy level difference is from n to the second energy level
So,
Initial energy level is ${n_i} = n$
Final energy level is ${n_f} = 2$
$\dfrac{1}{\lambda } = R(\dfrac{1}{4} - \dfrac{1}{{{n^2}}})$. . . . . (1)
From question $\lambda = \dfrac{{k{n^2}}}{{{n^2} - 4}}$. . . . . (2)
Equating (1) and (2)
$ \lambda = \dfrac{{k{n^2}}}{{{n^2} - 4}} = \dfrac{{4{n^2}}}{{R({n^2} - 4)}}$
$ \Rightarrow k = \dfrac{4}{R} $
Hence option (D) $k = \dfrac{4}{R}$ is the correct option
Additional information The Balmer series, or Balmer lines in atomic physics, is one of a set of six named series describing the spectral line emissions of the hydrogen atom. Spectral line emissions take place because an electron is moved from a higher energy level to a lower energy level the difference in energy is released as spectral line emissions.
Note: Often a mistake is made while taking the initial and final energy levels. For questions like these, the energy levels of the different spectrums are required to be known. Students might also make a mistake while taking the initial and final energy levels. An energy spectrum is released when an electron is moved from a higher energy level to a lower energy level. Hence the final energy level is always lower than the initial.
Formula used
Rydberg’s wavelength equation $\dfrac{1}{\lambda } = R(\dfrac{1}{{{n_f}}} - \dfrac{1}{{{n_i}}})$
Here,
Wavelength is represented by $\lambda $
Rydberg’s constant is represented by $R$
Initial and final energy levels are represented by ${n_i},{n_f}$ respectively
Complete step by step solution:
Rydberg’s equation for visible radiation of the hydrogen spectrum is given as
$\dfrac{1}{\lambda } = R(\dfrac{1}{{{n_f}^2}} - \dfrac{1}{{{n_i}^2}})$
For the Balmer series, the energy level difference is from n to the second energy level
So,
Initial energy level is ${n_i} = n$
Final energy level is ${n_f} = 2$
$\dfrac{1}{\lambda } = R(\dfrac{1}{4} - \dfrac{1}{{{n^2}}})$. . . . . (1)
From question $\lambda = \dfrac{{k{n^2}}}{{{n^2} - 4}}$. . . . . (2)
Equating (1) and (2)
$ \lambda = \dfrac{{k{n^2}}}{{{n^2} - 4}} = \dfrac{{4{n^2}}}{{R({n^2} - 4)}}$
$ \Rightarrow k = \dfrac{4}{R} $
Hence option (D) $k = \dfrac{4}{R}$ is the correct option
Additional information The Balmer series, or Balmer lines in atomic physics, is one of a set of six named series describing the spectral line emissions of the hydrogen atom. Spectral line emissions take place because an electron is moved from a higher energy level to a lower energy level the difference in energy is released as spectral line emissions.
Note: Often a mistake is made while taking the initial and final energy levels. For questions like these, the energy levels of the different spectrums are required to be known. Students might also make a mistake while taking the initial and final energy levels. An energy spectrum is released when an electron is moved from a higher energy level to a lower energy level. Hence the final energy level is always lower than the initial.
Recently Updated Pages
Difference Between Vapor and Gas: JEE Main 2024

Area of an Octagon Formula - Explanation, and FAQs

Charle's Law Formula - Definition, Derivation and Solved Examples

Central Angle of a Circle Formula - Definition, Theorem and FAQs

Average Force Formula - Magnitude, Solved Examples and FAQs

Boyles Law Formula - Boyles Law Equation | Examples & Definitions

Trending doubts
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

Degree of Dissociation and Its Formula With Solved Example for JEE

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Ideal and Non-Ideal Solutions Raoult's Law - JEE

Charging and Discharging of Capacitor

Electrical Field of Charged Spherical Shell - JEE

Other Pages
Displacement-Time Graph and Velocity-Time Graph for JEE

Two identical spheres carrying charges 9 uC and 5 uC class 12 physics JEE_Main

The average and RMS value of voltage for square waves class 12 physics JEE_Main

Explain the construction and working of a GeigerMuller class 12 physics JEE_Main

JEE Advanced Live Classes for 2025 By Vedantu

The electric field due to electric potential V 2x2 class 12 physics JEE_Main
