Rydberg Formula

Rydberg formula is a fundamental concept in atomic physics, providing insights into the behavior of electrons within atoms and the emission or absorption of light by these atoms. It offers a predictive tool for determining the wavelengths of light emitted or absorbed when electrons transition between different energy levels within an atom.The study of these spectral patterns, known as spectroscopy, originated from the observations made through the Rydberg formula. By understanding the relationship between the wavelengths of light and the energy levels of electrons within atoms, scientists can unravel the intricacies of atomic structure and behavior.

In this article, we will look deeper into the Rydberg formula, exploring its significance in atomic physics and its applications in spectroscopic analysis.

What is Rydberg Formula?

Rydberg formula, coined in honor of the Swedish physicist Johannes Rydberg, stands as a pivotal mathematical tool in the realm of atomic physics. This formula serves as a bridge between the wavelengths of spectral lines emitted or absorbed by hydrogen atoms and the energy levels governing the behavior of its electrons.The formula is primarily applicable to hydrogen-like atoms, which have only one electron. It paved the way for understanding the structure of atoms and the behavior of electrons within them. The Rydberg formula is instrumental in spectroscopy, the study of the interaction between matter and electromagnetic radiation.

𝑣ˉ  = 1/λ = R(1/n₁² – 1/n₂²)

Where,

• n₁ and n₂ are integers (n₁ < n₂),
• λ = Wavelength of photon,
• R = Rydberg constant [1.0973731568539(55) × 10⁷ m^-1].

We’ll deal with hydrogen in most cases, which allows us to employ the formula:

𝑣ˉ  = 1/λ = RH(1/n_1² – 1/n_2²)

Where,

• n₁ and n₂ are integers (n1 < n2),
• λ = Wavelength of photon,
• RH = Rydberg constant due to hydrogen [1.0973731568539(55) × 10⁷ m^-1].
  

Derivation of Rydberg Formula

When an electron transitions from one orbit to another,

ΔE = Ef – Ei

Where,

ΔE = Energy Difference,

Ef = Final Energy,

Ei = Initial Energy.

By using Bohr’s Model,

ΔE = (-RH/n_f²) – (-RH/n_i²)

∴ ΔE = RH/n_i² – RH/n_f²

∴ ΔE = RH(1/n_i² – 1/n_f²)

∴ ΔE = 2.18 × 10^-18(1/n_i² – 1/n_f²) ⇢ (Equation 1)

E = h{v}

Put the value of E in equation 1, and they get

∴ hv = 2.18 × 10^-18(1/n_i² – 1/n_f²)

v = 2.18 × 10-¹⁸/h × (1/n_i² – 1/n_f²) ⇢ (Where h = 6.626 × 10^-34)

v = 3.29 × 10¹⁵(1/n_i² – 1/n_f²) ⇢ (Equation 2)

They have, c = λ{v}

1/λ = v/c

Divide equation 2 by c,

v/c = 3.29 × 10¹⁵/c × (1/n_i² – 1/n_f²)

v¯ = 1/λ = 1.0974 × 10⁷(1/n_i² – 1/n_f²) m^-1

Solved Examples on Rydberg Formula

Example 1: Determine the wavelength of electromagnetic radiation emitted as an electron transitions from n=7 to n=3.

Solution:

Given: 

n₁ =3, n₂ =7, RH=1.0974×10⁷m⁻¹

Using the Rydberg formula:

1/𝜆=RH(1/n₁²−1/n₂²)

1/𝜆=1.0974×10⁷(1/9−1/49)

1/𝜆=1.0974×10⁷×(0.1111−0.0204)

1/𝜆=9.948×10⁵

λ≈100.5nm

Therefore, the wavelength of the electromagnetic radiation emitted is approximately λ≈100.5nm

Example 2: Calculate the wavelength of electromagnetic radiation emitted by an electron transitioning from n=12 to n=5.

Solution: 

Given: 

n₁ =5, n2 =12, RH=1.0974×10⁷m−1

Using the Rydberg formula:

1/𝜆=RH(1/n₁²−1/n₂²)

1/𝜆=1.0974×10⁷(1/25−1/144)

1/𝜆=1.0974×10⁷×(0.04−0.0069)

1/𝜆=3.635×10⁵

λ≈2752nm

Example 3: Calculate the wavelength of electromagnetic radiation emitted by an electron transitioning from n=8 to n=2.

Solution:

Given: 

n₁ =2, n₂ =8, RH=1.0974×10⁷m⁻¹

Using the Rydberg formula:

1/𝜆=RH(1/n₁²−1/n₂²)

1/𝜆=1.0974×10⁷(1/4−1/64)

1/𝜆=1.0974×10⁷×(0.25−0.015625)

λ≈3.88×10⁻⁷ m