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Bohr's Theory of Hydrogen-like Atoms


Bohr's theory of the hydrogen atom is based on the idea that electrons orbit the nucleus in specific energy levels or shells, and that they can only occupy certain quantized energy states.

According to Bohr's theory, the electron can only occupy specific energy levels, or shells, which are determined by the electron's energy. When the electron moves from a higher energy level to a lower one, it emits a photon of electromagnetic radiation with a specific frequency.

bohr's model

The energy in nth Bohr's orbit is given by \begin{align} E_n & =-\frac{mZ^2e^4}{8{\epsilon_0}^2 h^2 n^2},\\ E_n & =-\frac{13.6 Z^2}{n^2}\;\mathrm{eV} \end{align}

Bohr's theory also introduced the concept of the Bohr radius, which is the average distance between the electron and the nucleus in the ground state of the hydrogen atom. This radius is a fundamental constant in atomic physics and is equal to 0.529 angstroms. The radius of nth Bohr's orbit is given by \begin{align} r_n & =\frac{\epsilon_0 h^2 n^2}{\pi m Z e^2}, \\ r_n &=\frac{n^2a_0}{Z},\\ a_0 &=0.529\;\mathrm{Angstrom} \end{align}

The angular momentum is quantized and its value in nth Bohr's orbit is given by \begin{align} l=\frac{nh}{2\pi} \end{align}

The energy required to move an electron from one energy level to another can be calculated using a formula that relates the difference in energy levels to the frequency of the emitted photon.

state transition

The photon energy in state transition is given by \begin{align} E_2-E_1=h\nu \end{align}

The wavelength of emitted radiation for a transition from $n$th to $m$th state is given by \begin{align} \frac{1}{\lambda}=RZ^2 \left[\frac{1}{n^2}-\frac{1}{m^2}\right] \end{align}

Problems from IIT JEE

Problem (IIT JEE 2011): The wavelength of the first spectral line in the Balmer series of hydrogen atom is ${6561}\;{Angstrom}$. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is

  1. ${1215}\;{Angstrom}$
  2. ${1640}\;{Angstrom}$
  3. ${2430}\;{Angstrom}$
  4. ${4687}\;{Angstrom}$
The wavelength of spectral line when an electron in singly ionized atom of atomic number $Z$ makes a transition from $m^\text{th}$ to $n^\text{th}$ orbit ($m>n$) is, \begin{align} \frac{1}{\lambda}=RZ^2\left[\frac{1}{n^2}-\frac{1}{m^2}\right]. \end{align} In the Balmer series, first spectral line corresponds to transition from $m=3$ to $n=1$ and second spectral line corresponds to transition from $m=4$ to $n=2$. Thus, for hydrogen ($Z=1$), \begin{align} \frac{1}{6561} & =R\left[\frac{1}{2^2}-\frac{1}{3^2}\right] \\ &=\frac{5R}{36}, \end{align} and for singly ionized helium ($Z=2$), \begin{align} \frac{1}{\lambda} &=R\times 2^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right]\\ &=\frac{3R}{4}. \end{align} Divide above equations to get $\lambda={1215}\;{Angstrom}$.


  1. Rutherford's Model of Atom
  2. Characteristic and Continuous X-rays New
  3. Moseley's Law
JEE Physics Solved Problems in Mechanics