Hydrogen Atom Spectral Series, Excitation Energy and Ionisation Energy: Concepts, Potentials and Solved Numericals

Class 12/Modern Physics/NEET & JEE Study Materials

Introduction

The hydrogen atom is the simplest atom, consisting of one proton in its nucleus and one electron revolving around it. The study of its energy levels and spectral lines played a major role in the development of atomic physics.

According to Niels Bohr’s atomic model, the electron in a hydrogen atom can occupy only certain permitted energy levels. When the electron moves from one energy level to another, it absorbs or emits energy in the form of electromagnetic radiation.

In this article, we will study the spectral series of hydrogen, excitation energy, ionisation energy, excitation potential, ionisation potential, and important solved numerical problems for CBSE Class 12 Physics, NEET, and JEE.

1. Energy Levels of the Hydrogen Atom

According to Bohr’s theory, the energy of an electron in the nth permitted orbit of a hydrogen atom is given by:

Eₙ = −13.6/n² eV

where:

• Eₙ is the energy of the electron in the nth orbit.
• n is the principal quantum number (1, 2, 3, …).
• eV stands for electron volt, a unit of energy.

The energy of the electron in the ground state (n = 1) is −13.6 eV.

Energy level	n	Energy
Ground state	1	−13.6 eV
First excited state	2	−3.40 eV
Second excited state	3	−1.51 eV
Third excited state	4	−0.85 eV
Fifth level	5	−0.544 eV
Ionisation limit	∞	0 eV

The negative sign indicates that the electron is bound to the nucleus. At n = ∞, the energy becomes zero, representing an electron just free from the atom.

2. Spectral Series of the Hydrogen Atom

When an electron transitions from a higher energy level to a lower energy level, it emits a photon. The collection of spectral lines produced by transitions ending at a particular lower energy level is called a spectral series.

The energy of the emitted photon is:

Photon energy = Eᵢ − E𝒇

Here, Eᵢ is the initial energy and E𝒇 is the final energy, with Eᵢ corresponding to the higher level.

The wavelength of the emitted radiation can be calculated using the Rydberg formula:

1/λ = R × (1/n𝒇² − 1/nᵢ²)

where:

• λ is the wavelength of the emitted radiation.
• R is the Rydberg constant, approximately 1.097 × 10⁷ m⁻¹.
• n𝒇 is the lower energy level.
• nᵢ is the higher energy level, where nᵢ > n𝒇.

Important spectral series

Spectral series	Final energy level	Region
Lyman series	n = 1	Ultraviolet
Balmer series	n = 2	Mainly visible, with some lines near ultraviolet
Paschen series	n = 3	Infrared
Brackett series	n = 4	Infrared
Pfund series	n = 5	Infrared
Humphreys series	n = 6	Infrared

Lyman series

All transitions end at n = 1.

Examples: n = 2 → 1, n = 3 → 1, n = 4 → 1.

These transitions produce ultraviolet radiation.

Balmer series

All transitions end at n = 2.

Examples: n = 3 → 2, n = 4 → 2, n = 5 → 2.

The Balmer series includes the prominent visible hydrogen lines.

Paschen, Brackett, Pfund and Humphreys series

These series are produced when electrons transition to n = 3, 4, 5 and 6, respectively. They lie in the infrared region.

Important point: As the initial energy level increases, the spectral lines in a series become closer together and converge towards a series limit.

3. Excitation Energy

Excitation energy is the minimum energy required to raise an electron from its ground state to a specified higher energy level without removing it from the atom.

For hydrogen, the excitation energy from the ground state to the nth level is:

Excitation energy = Eₙ − E₁

Substituting the Bohr energy values:

Excitation energy = 13.6 × (1 − 1/n²) eV

For the first excited state, n = 2:

Excitation energy = 13.6 × (1 − 1/4)

Excitation energy = 10.2 eV

For the second excited state, n = 3:

Excitation energy = 13.6 × (1 − 1/9)

Excitation energy ≈ 12.09 eV

The excitation energy increases as the electron is raised to higher energy levels, approaching 13.6 eV as n approaches infinity.

4. Excitation Potential

Excitation potential is the potential difference through which an electron must be accelerated to gain sufficient energy to excite a hydrogen atom from its ground state to a specified higher energy level.

Since an electron gains 1 eV of energy when accelerated through a potential difference of 1 V, the numerical value of the excitation potential in volts equals the excitation energy in electron volts.

Excitation potential (in volts) = Excitation energy (in eV)

Examples:

• Excitation potential for n = 2: 10.2 V.
• Excitation potential for n = 3: approximately 12.09 V.
• Excitation potential for n = 4: 12.75 V.

These values refer to excitation from the ground state.

5. Ionisation Energy

Ionisation energy is the minimum energy required to remove an electron completely from an isolated atom in a specified energy state.

For hydrogen in its ground state, the electron energy is −13.6 eV. At the ionisation limit, the energy is zero.

Therefore:

Ionisation energy = 0 − (−13.6)

Ionisation energy = 13.6 eV

To convert electron volts into joules, use:

1 eV = 1.602 × 10⁻¹⁹ J

Hence:

13.6 eV ≈ 2.18 × 10⁻¹⁸ J

For hydrogen, the ionisation energy from the nth level is:

Ionisation energy from nth level = 13.6/n² eV

Examples:

• From n = 1: 13.6 eV.
• From n = 2: 3.40 eV.
• From n = 3: approximately 1.51 eV.
• From n = 4: 0.85 eV.

The higher the initial energy level, the less energy is needed to remove the electron.

6. Ionisation Potential

Ionisation potential is the minimum potential difference through which an electron must be accelerated to supply the energy required for ionisation, under the usual introductory atomic-physics definition.

For a hydrogen atom in its ground state:

Ionisation potential = 13.6 V

For an electron in the nth energy level:

Ionisation potential = 13.6/n² V

For example, the ionisation potential from n = 2 is 3.40 V.

Remember: Ionisation energy is measured in eV or joules, whereas ionisation potential is measured in volts. Their numerical values are equal when energy is expressed in eV and the corresponding potential is expressed in volts.

7. Difference Between Excitation and Ionisation

Excitation	Ionisation
The electron moves to a higher bound energy level.	The electron is removed from the atom.
The final energy remains negative for a bound state.	The final energy reaches zero or above, using the isolated-atom reference.
Excitation energy from the ground state is less than 13.6 eV for a finite excited level.	Ground-state ionisation energy is 13.6 eV.
The electron remains bound to the nucleus.	The electron becomes free from the atom.

8. Important Formulas for Numerical Problems

1. Energy of the nth level: Eₙ = −13.6/n² eV.
2. Excitation energy from the ground state: 13.6 × (1 − 1/n²) eV.
3. Ionisation energy from the nth level: 13.6/n² eV.
4. Photon energy: E = hν = hc/λ.
5. Rydberg formula: 1/λ = R × (1/n𝒇² − 1/nᵢ²).
6. Conversion: 1 eV = 1.602 × 10⁻¹⁹ J.
7. For hydrogen, excitation potential in volts has the same numerical value as excitation energy in eV.
8. For hydrogen, ionisation potential in volts has the same numerical value as ionisation energy in eV.

Useful constants:

• Rydberg constant, R = 1.097 × 10⁷ m⁻¹.
• Planck constant, h = 6.626 × 10⁻³⁴ J s.
• Speed of light, c = 3.0 × 10⁸ m/s.
• 1 eV = 1.602 × 10⁻¹⁹ J.

9. Solved Numerical Problems

Numerical 1: Excitation energy to the first excited state

Question: Calculate the energy required to excite a hydrogen atom from n = 1 to n = 2.

Solution:

Energy at n = 1 = −13.6 eV.

Energy at n = 2 = −13.6/4 = −3.4 eV.

Excitation energy = −3.4 − (−13.6)

Answer: 10.2 eV

Numerical 2: Excitation energy to the third energy level

Question: Find the excitation energy required to raise an electron in hydrogen from the ground state to n = 3.

Solution:

E₁ = −13.6 eV.

E₃ = −13.6/9 = −1.51 eV approximately.

Excitation energy = E₃ − E₁

= −1.51 − (−13.6)

Answer: 12.09 eV approximately

Numerical 3: Ionisation energy from the second energy level

Question: Calculate the ionisation energy of a hydrogen atom when its electron is in n = 2.

Solution:

Ionisation energy = 13.6/n² eV

= 13.6/2²

= 13.6/4

Answer: 3.40 eV

Numerical 4: Convert hydrogen ionisation energy into joules

Question: Express the ground-state ionisation energy of hydrogen in joules.

Solution:

Ionisation energy = 13.6 eV.

Using 1 eV = 1.602 × 10⁻¹⁹ J:

Energy = 13.6 × 1.602 × 10⁻¹⁹

Answer: 2.18 × 10⁻¹⁸ J approximately

Numerical 5: Wavelength of the first Balmer line

Question: Calculate the wavelength of radiation emitted when an electron transitions from n = 3 to n = 2.

Solution:

Using the Rydberg formula:

1/λ = R × (1/2² − 1/3²)

= 1.097 × 10⁷ × (1/4 − 1/9)

= 1.097 × 10⁷ × 5/36

Therefore, λ ≈ 6.563 × 10⁻⁷ m.

Converting metres into nanometres:

λ ≈ 656.3 nm.

Answer: 656.3 nm approximately. This is the red H-alpha line of hydrogen.

Numerical 6: Wavelength of the first Lyman line

Question: Calculate the wavelength emitted when an electron falls from n = 2 to n = 1.

Solution:

1/λ = R × (1/1² − 1/2²)

= 1.097 × 10⁷ × (1 − 1/4)

= 1.097 × 10⁷ × 3/4

Therefore, λ ≈ 1.216 × 10⁻⁷ m.

Answer: 121.6 nm approximately, which lies in the ultraviolet region.

Numerical 7: Excitation potential for n = 4

Question: Calculate the excitation potential required to raise a hydrogen atom from n = 1 to n = 4.

Solution:

Excitation energy = 13.6 × (1 − 1/4²) eV

= 13.6 × (15/16)

= 12.75 eV.

Therefore, the excitation potential is 12.75 V.

Answer: 12.75 V

Numerical 8: Ionisation potential from n = 3

Question: Find the ionisation potential of hydrogen when the electron is in the third energy level.

Solution:

Ionisation potential = 13.6/n² V

= 13.6/9

Answer: approximately 1.51 V

Numerical 9: Energy of a photon emitted in a transition

Question: Calculate the energy emitted when an electron falls from n = 3 to n = 2 in hydrogen.

Solution:

E₃ = −13.6/9 = −1.51 eV approximately.

E₂ = −13.6/4 = −3.40 eV.

Photon energy = E₃ − E₂

= −1.51 − (−3.40)

Answer: approximately 1.89 eV

Numerical 10: Minimum wavelength in the Balmer series

Question: Calculate the limiting wavelength of the Balmer series.

Solution:

For the Balmer series, n𝒇 = 2. At the series limit, nᵢ approaches infinity.

1/λ = R × (1/2² − 1/∞²)

Since 1/∞² = 0:

1/λ = R/4

λ = 4/R

= 4/(1.097 × 10⁷)

Answer: approximately 364.6 nm

This is the short-wavelength limit of the Balmer series.

10. Practice Questions

Try solving these questions before checking the answers.

1. Calculate the excitation energy required to raise a hydrogen electron from n = 1 to n = 4.
2. Find the ionisation energy of hydrogen when the electron occupies n = 3.
3. Calculate the excitation potential for the transition from the ground state to n = 2.
4. Identify the spectral series for a transition from n = 5 to n = 1.
5. Calculate the energy of the photon emitted when an electron falls from n = 4 to n = 2.
6. Find the ionisation energy in joules for a hydrogen electron in n = 2.
7. What is the series limit of the Lyman series?
8. Distinguish between excitation potential and ionisation potential.

Answers

1. 12.75 eV.
2. Approximately 1.51 eV.
3. 10.2 V.
4. Lyman series.
5. 2.55 eV.
6. Approximately 5.45 × 10⁻¹⁹ J.
7. Approximately 91.2 nm.
8. Excitation potential raises an electron to a higher bound level; ionisation potential supplies the energy needed to remove it from the atom.

11. Common Mistakes to Avoid

• Forgetting the negative sign in the energy of a bound electron.
• Using the higher energy level as the final level in the Rydberg formula for emission.
• Confusing excitation energy with ionisation energy.
• Writing energy units as volts or potential units as electron volts.
• Assuming every hydrogen spectral line lies in the visible region.
• Forgetting that ionisation energy depends on the initial energy level of the electron.

Conclusion

The hydrogen atom provides a fundamental example of quantised energy levels. Its spectral series arise from electronic transitions between permitted energy states. Excitation energy and excitation potential describe the energy needed to raise an electron to a higher bound state, while ionisation energy and ionisation potential describe the energy required to remove the electron from the atom.

Understanding these concepts, remembering the Bohr energy formula and Rydberg formula, and practising numerical problems will help students strengthen their preparation for CBSE Class 12 Physics, NEET, and JEE examinations.

Study smart with eduPhysics CBSE AND NEET: Revise the formulas, practise the solved examples, and test your understanding with additional MCQs and previous-year questions.

https://edu-physics.com/2021/02/05/bohrs-model-hydrogen-atom-postulates-radius-energy/