Bohr Model Hydrogen Atom: Energy Transitions from Balmer Series to Lyman Series

By RCM, IITian · October 13, 2026 · Updated October 19, 2026 · 43 min study · Physics

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Bohr Model Hydrogen Atom is a very important topic in Modern Physics. In this article we discuss and derive the equations for cahnge in kinetic energy, total energy, de Broglie wavelength, and the frequency of the emitted photon when an electron undergoes a transition from the lowest energy state of Balmer Series to the lowest energy sate of the Lyman series. We will also solve a practice problem on this energy transitions in Bohr Model Hydrogen Atom. Here we use the principle of quantization of angular momentum and centripetal force in circular orbits.

Problem Based on Bohr Model Hydrogen Atom

Question:

Find the following when an electron undergoes a transition from the lowest energy state in Balmer Series to the lowest energy state in Lyman Series in Bohr Model Hydrogen Atom.

Magnitude of change in electron’s Kinetic Energy
The frequency of emitted radiation
The magnitude of change in electron’s total energy
The magnitude of change if electron’s de Broglie wavelength

Solution:

The above problem can be divided into two parts.

Deriving the governing equations in Bohr Model Hydrogen Atom
Solving problem by substituting numerical values.

Derivations of Equations for Bohr Model Hydrogen Atom for energy transitions

Finding initial and final energy states

Here to solve the problem of Bohr Model Hydrogen Atom, we are required to find the Initial and final energy states of the electron. It is the most important concept to be understood to solve the given problem.

Let us try to understand the energy levels of Balmer Series and Lyman Series. The lowest energy state in Balmer Series is at n=3 and the lowest energy state in Lyman series is at n=1. Hence the transition from lowest energy state in Balmer series to the lowest energy state in Lyman series is from n=3 to n=1.

The below image explains clearly the energy states required to be considered for solving above problem.

Bohr Model Hydrogen Atom- Balmer Series, Lyman Series

The red arrow line shows the energy transition levels to be considered in the Bohr Model Hydrogen Atom when the subject electron jumps from the lowest energy level in Balmer series to lowest Energy level in Lyman Series.

Now let us try to derive the required equations for the asked parameters in the problem.

Derivation of cahnge in Electron’s Kinetic Energy

The above can be found out using the quantization of angular momentum principle and calculation of centripetal force in circular orbit.

Quantization of Angular Momentum

For an electron in any $k^{th} orbit$ , as per Bohr Model Hydrogen Atom, from quantization of angular momentum,

${mv}_{k} r_{k} = \frac{kh}{2 π}$

Electrostatic centripetal force for circular orbit

For an electron in any $k^{th} orbit$ , as per Bohr Model Hydrogen Atom, from electrostatic forces and centripetal forces,

$\frac{{mv}_{k}^{2}}{r_{k}} = \frac{1}{4 {π ε}_{0}} \frac{e^{2}}{r_{k}^{2}}$

Kinetic Energy calculations

For an electron in any $k^{th} orbit$ , as per Bohr Model Hydrogen Atom, Kinetic Energy calculated as below,

$K_{k} = \frac{1}{2} {mv}_{k}^{2}$ , but,

${mv}_{k} r_{k} = \frac{kh}{2 π}$ , hence,

$m = \frac{kh}{2 π v_{k} r_{k}}$

By substituting this value of m in the Kinetic Energy Equation , we get,

$K_{k} = \frac{1}{2} {mv}_{k}^{2} = \frac{1}{2} (\frac{kh}{2 π v_{k} r_{k}}) v_{k}^{2} = \frac{{khv}_{k}}{4 π r_{k}}$

Kinetic Energy in $K^{th} Orbit$ is given by,

1. Magnitude of Change in Electron’s Kinetic Energy

For the given problem, change in Kinetic Energy from n=3 to n=1 is given by,

2. Magnitude of frequency of emitted radiation from electron due to energy transition

By combining the Force (Electrostatic forces & Centripetal Force) Equations and Kinetic Energy Equations, From

\frac{{mv}_{k}^{2}}{r_{k}} = \frac{1}{4 {π ε}_{0}} \frac{e^{2}}{r_{k}^{2}}

and

K_{k} = \frac{1}{2} {mv}_{k}^{2}

, we get,

$K_{k} = \frac{1}{2} {mv}_{k}^{2} = \frac{1}{2} (r_{k} (\frac{1}{4 {π ε}_{0}} \frac{e^{2}}{r_{k}^{2}})) = \frac{e^{2}}{8 {π ε}_{0} r_{k}}$

Potential Energy is given by,

$U_{k} = - \frac{e^{2}}{4 {π ε}_{0} r_{k}}$

Total Energy E = $K_{k} + U_{k} = - \frac{e^{2}}{8 {π ε}_{0} r_{k}}$

Planck-Einstein Relation

From Planck-Einstein equation ,

Total Energy =E = $h υ$ , and,

Total Energy E = $K_{k} + U_{k} = - \frac{e^{2}}{8 {π ε}_{0} r_{k}}$ ,

Combining above two equations , we get,

$Frequency of emitted radiation = υ = \frac{e^{2}}{8 {π ε}_{0} h} (\frac{1}{r_{1}} - \frac{1}{r_{3}})$

3. The magnitude of change in electron’s total energy

The magnitude of Change in total energy = K+U. But from above discussions, in Bohr Model Hydrogen Atom, U=-2K. Hence, Total Energy = E= K+(-2K) = -K The magnitude of change in Total energy is same as the the magnitude of the change in Kinetic Energy if the absolute values are taken. Hence, Change in Total Energy is given by the Equation,

4. The magnitude of change in electron’s de Broglie wavelength

In Bohr Model Hydrogen Atom, the de Broglie wavelength is given by,

λ_{k} = \frac{h}{{mv}_{k}} = \frac{h}{(\frac{kh}{2 π r_{k}})} = \frac{2 π r_{k}}{k} = \frac{2 π a_{0} k^{2}}{k} = 2 π a_{0} k

$| Δ λ | = | λ_{3} - λ_{1} | = 2 π a_{0} | 3 - 1 | = 4 π a_{0}$

Where $a_{0} = 5.29 \times 10^{- 11} m$

Hence the magnitude of change in de Broglie wavelength is given by the formula,

As above the equations for all 4 quantities derived while discussing the problem involving Bohr Model Hydrogen Atom.

Numerical Solution:

By substituting the values of Planck constant, orbit numbers, electron charge, we get the answers as below:

$Magnitude of change in Kinetic Energy = 12.09 eV$
$Frequency of emitted Photon = 2.924 \times 10^{15} Hz$
$Magnitude of change in Total Energy = 12.09 eV$
Magnitude of change in de Broglie wavelength = $6.65 \times 10^{- 10} m$

The detailed solution and numerical answers are given as above. This is a very important topic and full or part of questions being asked in various examinations.

Advice

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