Class 12/Modern Physics/NEET and JEE Preparation
Introduction
The structure of the atom was one of the major challenges in the development of modern physics. Rutherford’s nuclear model established that an atom contains a small, positively charged nucleus surrounded by electrons. However, it could not explain why electrons do not continuously lose energy and fall into the nucleus, according to classical electromagnetic theory.
In 1913, Danish physicist Niels Bohr proposed a model of the hydrogen atom that successfully explained its stability and line spectrum. Bohr introduced the revolutionary idea that electrons can occupy only certain permitted orbits with definite energies.
This article explains the postulates of Bohr’s atomic model, the quantisation of angular momentum, the radii of permitted electron orbits and the energy of electrons in the hydrogen atom. These concepts are essential for CBSE Class 12 Physics, NEET and JEE examinations.
1. Postulates of Bohr’s Theory of the Hydrogen Atom
1. Postulates of Bohr’s Theory of the Hydrogen Atom
Bohr’s model is based on the following fundamental postulates.
Postulate 1: Electrons revolve in permitted stationary orbits
An electron revolves around the positively charged nucleus in certain permitted circular orbits called stationary orbits.
While moving in a stationary orbit, the electron does not radiate energy, even though it is accelerating towards the nucleus.
Each permitted orbit corresponds to a definite energy of the electron.
Postulate 2: Angular momentum is quantised
Only those orbits are permitted for which the angular momentum of the electron is an integral multiple of the reduced Planck constant.
Bohr’s quantisation condition:
mvr = nh/(2π)
Here:
- m = mass of the electron
- v = speed of the electron
- r = radius of the orbit
- h = Planck’s constant
- n = principal quantum number, where n = 1, 2, 3, …
The principal quantum number identifies the permitted orbit. The first orbit corresponds to n = 1, the second to n = 2, and so on.
Postulate 3: Radiation is emitted or absorbed during transitions
An electron emits or absorbs electromagnetic radiation when it transitions between two permitted energy levels.
The energy of the emitted or absorbed photon equals the difference between the energies of the two states.
Photon energy = |E₂ − E₁| = hν
Here, ν is the frequency of the radiation.
When an electron moves from a higher energy level to a lower energy level, a photon is emitted. When it moves from a lower energy level to a higher energy level, a photon is absorbed.
Postulate 4: The lowest energy orbit is the ground state
The orbit with n = 1 has the lowest energy and is called the ground state. Orbits with n = 2, 3, 4, and higher values represent excited states.
An electron requires energy to move from the ground state to an excited state.
2. Bohr’s Quantisation Condition of Angular Momentum
According to classical mechanics, an electron moving in a circular orbit has angular momentum given by:
Angular momentum, L = mvr
Bohr proposed that the angular momentum of an electron cannot have an arbitrary value. It must satisfy the quantisation condition:
L = nh/(2π) = nℏ
where ℏ = h/(2π).
Therefore, the permitted angular momentum values are:
- First orbit: L = ℏ
- Second orbit: L = 2ℏ
- Third orbit: L = 3ℏ
- Fourth orbit: L = 4ℏ
The angular momentum is quantised because only discrete values are allowed.
Derivation of the quantisation condition
For an electron of mass m and speed v revolving around a nucleus of charge +Ze, the electrostatic force provides the centripetal force.
For a hydrogen-like atom, the force equation is:
mv²/r = (1/4πε₀)(Ze²/r²)
Rearranging:
mv² = Ze²/(4πε₀r)
Bohr’s quantisation condition is:
mvr = nℏ
Squaring this equation gives:
m²v²r² = n²ℏ²
Using the force equation to eliminate v, we obtain the radius of the nth permitted orbit:
rₙ = (4πε₀ℏ²/mZe²)n²
For hydrogen, Z = 1. Hence, the orbit radius is proportional to n².
Important: Bohr’s quantisation condition is a fundamental postulate of his original atomic model. It should not be interpreted as a complete description of electron motion in modern quantum mechanics.
3. Radii of Permitted Orbits in the Hydrogen Atom
The radius of a permitted orbit is obtained by combining the electrostatic force equation with Bohr’s angular momentum quantisation condition.
For a hydrogen atom:
rₙ = a₀n²
where a₀ is the Bohr radius.
The Bohr radius is approximately:
a₀ = 0.529 Å = 5.29 × 10⁻¹¹ m
Therefore:
- First orbit, n = 1: r₁ = a₀
- Second orbit, n = 2: r₂ = 4a₀
- Third orbit, n = 3: r₃ = 9a₀
- Fourth orbit, n = 4: r₄ = 16a₀
The radius of the nth orbit increases as the square of the principal quantum number.
Important observation
If the radius of the first Bohr orbit is r₁, then:
rₙ/r₁ = n²
For example, the radius of the third orbit is nine times the radius of the first orbit.
For a hydrogen-like ion with nuclear charge +Ze, the corresponding expression is:
rₙ = a₀n²/Z
Thus, for the same principal quantum number, the orbit radius decreases as the nuclear charge increases.
4. Energy of an Electron in the Hydrogen Atom
The total energy of an electron in a permitted orbit is the sum of its kinetic energy and electrostatic potential energy.
Total energy, E = K + U
Kinetic energy
The electrostatic force provides the centripetal force:
mv²/r = e²/(4πε₀r²)
Multiplying by r/2 gives:
(1/2)mv² = e²/(8πε₀r)
Therefore, the kinetic energy is:
K = e²/(8πε₀r)
The kinetic energy is positive.
Potential energy
Taking the electrostatic potential energy to be zero when the electron and proton are infinitely far apart, the potential energy of the electron-proton system is:
U = −e²/(4πε₀r)
The potential energy is negative because the electron and proton attract each other.
Total energy
Adding the kinetic and potential energies:
E = K + U
E = e²/(8πε₀r) − e²/(4πε₀r)
Therefore:
E = −e²/(8πε₀r)
Substituting the radius of the nth Bohr orbit gives the energy of the electron in the hydrogen atom.
Energy of the nth orbit:
Eₙ = −13.6/n² eV
Equivalently:
Eₙ = −13.6/n² eV, where n = 1, 2, 3, …
The negative sign indicates that the electron is bound to the nucleus. Energy must be supplied to remove it completely from the atom.
Energy levels of hydrogen
| Principal quantum number (n) | Orbit | Energy |
| 1 | First orbit | −13.6 eV |
| 2 | Second orbit | −3.40 eV |
| 3 | Third orbit | −1.51 eV |
| 4 | Fourth orbit | −0.85 eV |
| ∞ | Ionisation limit | 0 eV |
As n increases, the energy becomes less negative and approaches zero. The energy levels become increasingly close together at higher values of n.
5. Ionisation Energy of the Hydrogen Atom
Ionisation energy is the minimum energy required to remove an electron completely from an atom in its ground state.
For hydrogen, the ground-state energy is −13.6 eV. At the ionisation limit, the energy is zero.
Therefore:
Ionisation energy = 0 − (−13.6 eV)
Ionisation energy of hydrogen = 13.6 eV
This value applies to an isolated hydrogen atom initially in its ground state.
If the electron is initially in the nth orbit, the energy required for ionisation is 13.6/n² eV.
6. Energy Difference Between Two Bohr Orbits
When an electron transitions between two energy levels, the energy of the photon is equal to the magnitude of the energy difference.
For a transition from an initial level nᵢ to a final level n𝒻:
Photon energy = |E𝒻 − Eᵢ|
For emission, nᵢ is greater than n𝒻. The electron moves to a lower energy level and emits a photon.
For absorption, n𝒻 is greater than nᵢ. The electron absorbs a photon to reach a higher energy level.
Since photon energy is hν:
hν = |E𝒻 − Eᵢ|
The corresponding wavelength can be found using:
λ = hc/|E𝒻 − Eᵢ|
This explains why hydrogen produces a line spectrum rather than a continuous spectrum. Only particular energy differences, and therefore particular photon frequencies, are permitted.
7. Important Formulas for CBSE, NEET and JEE
Revise the following formulas before solving numerical problems:
- Angular momentum: L = mvr
- Bohr’s quantisation condition: mvr = nh/(2π)
- Radius of the nth hydrogen orbit: rₙ = a₀n²
- Bohr radius: a₀ = 5.29 × 10⁻¹¹ m
- Energy of the nth hydrogen orbit: Eₙ = −13.6/n² eV
- Ground-state energy: E₁ = −13.6 eV
- Ionisation energy from the ground state: 13.6 eV
- Photon energy during a transition: hν = |E𝒻 − Eᵢ|
- Photon wavelength: λ = hc/|E𝒻 − Eᵢ|
For hydrogen-like ions, remember that the orbit radius varies as n²/Z and the energy varies as −Z²/n².
8. Conclusion
Bohr’s theory of the hydrogen atom was a major step towards understanding atomic structure. Its central ideas are the existence of stationary energy states, quantisation of angular momentum and the emission or absorption of photons during electronic transitions.
The radius of a permitted orbit increases as n², while the magnitude of the electron’s binding energy decreases as 1/n². These relationships are important for understanding atomic spectra, ionisation energy and electronic transitions.
Although modern quantum mechanics has replaced Bohr’s circular-orbit picture with a more complete description of electrons, Bohr’s model remains an essential foundation for studying atomic physics.Prepare for your next Physics examination with eduPhysics. Explore our CBSE Physics notes, practical manuals, NEET MCQ question banks and exam-oriented study materials at edu-physics.com.
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