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Rydberg Formula

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  • Last Updated : 01 May, 2022

The Rydberg formula predicts the wavelength of light produced by an electron moving between atomic energy levels. Each element has a unique spectral fingerprint. When an element’s gaseous state is heated, light is produced. When this light passes through a prism or a diffraction grating, bright lines of various hues can be discerned. In some manner, each element stands out from the rest. Spectroscopic studies began with this discovery. The Rydberg formula will be discussed in-depth in this article.

Rydberg 

The Rydberg formula is a mathematical formula for calculating the wavelength of light emitted by an electron moving between the energy levels of an atom. When an electron goes from one atomic orbital to the next, its energy fluctuates. When an electron shift from a high-energy orbital to a lower-energy orbital, a photon of light is created. A photon of light is absorbed by the atom when an electron moves from a low-energy to a higher-energy state. The Rydberg Formula can be used to compute various elements’ spectra.

Rydberg Formula

The Rydberg Formula is used to calculate the spectra of various elements and is written as

\bar{v}   = 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:

\bar{v}   = 1/λ = RH(1/n12 – 1/n22)

Where,

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

\bar{v}   = 1/λ = 1.0974 × 107(1/n12 – 1/n22)

Derivation of Rydberg Formula

When an electron moves 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/nf2) – (-RH/ni2)

∴ ΔE = RH/ni2 – RH/nf2

∴ ΔE = RH(1/ni2 – 1/nf2)

∴ ΔE = 2.18 × 10-18(1/ni2 – 1/nf2) ⇢ (Equation 1)

E = h{v}

Put value of E in equation 1, we get

∴ hv = 2.18 × 10-18(1/ni2 – 1/nf2)

v = 2.18 × 10-18/h × (1/ni2 – 1/nf2) ⇢ (Where, h = 6.626 × 10-34)

v = 3.29 × 1015(1/ni2 – 1/nf2) ⇢ (Equation 1)

We have, c = λ{v}

1/λ = v/c

Divide equation 2 by c,

v/c = 3.29 × 1015/c × (1/ni2 – 1/nf2)

\bar{v}   = 1/λ = 1.0974 × 107(1/ni2 – 1/nf2) m-1

Sample Questions

Question 1: Why does Rydberg solely use hydrogen in his experiments?

Answer:

The Rydberg equation can only be used with hydrogen and other hydrogenic compounds because it is an empirical formula based on the Bohr model of the hydrogen atom.
The Rydberg constant is used to calculate the wavelengths in the hydrogen spectrum, which are the amount of energy received or released as photons as electrons travel between shells in the hydrogen atom.

Question 2: What is the significance of the Rydberg constant?

Answer:

The Rydberg constant is one of the most important constants in atomic physics because of its link to the fundamental atomic constants (e, h, me and c) and the accuracy with which it can be calculated. The first publication of the Rydberg constant in scientific literature was in 1890. The Rydberg constant is used to calculate the wavelengths in the hydrogen spectrum, which are the amount of energy received or released as photons as electrons travel between shells in the hydrogen atom.
To put it another way, the Rydberg formula is a mathematical equation that predicts the wavelength of light produced by an electron traversing between energy levels in an atom. When an electron goes from one atomic orbital to the next, its energy fluctuates.

Question 3: Find the wavelength of an electron’s electromagnetic radiation as it relaxes from n = 7 to n = 3.

Solution:

Given: n1 = 3, n2 = 7, RH = 1.0974 × 107

Since,

1/λ = RH(1/n12 – 1/n22)

∴ 1/λ = 1.0974 × 107(1/32 – 1/72)

∴ 1/λ = 1.0974 × 107(1/9 – 1/49)

∴ 1/λ = 1.0974 × 107(0.1 – 0.02)

∴ 1/λ = 1.0974 × 107 × 0.08

∴ 1/λ = 0.0877 × 107

∴ λ = 11.402 × 10-7 m

Question 4: If an electron transition occurs from n1 = 2 to n2 = 3, what is the wavelength of the emitted photon?

Solution:

Given: n1 = 1, n2 = 3, RH = 1.0974 × 107

Since,

1/λ = RH(1/n12 – 1/n22)

∴ 1/λ = 1.0974 × 107(1/12 – 1/32)

∴ 1/λ = 1.0974 × 107(1/1 – 1/9)

∴ 1/λ = 1.0974 × 107(0.1666)

∴ 1/λ = 0.182 × 107

∴ λ = 5.47 × 10-7 m

Question 5: Determine the wavelength of electromagnetic radiation emitted by an electron as it relaxed from n = 12 to n = 5.

Solution:

Given: n1 = 5, n2 = 12, RH = 1.0974 × 107

Since,

1/λ = RH(1/n12 – 1/n22)

∴ 1/λ = 1.0974 × 107(1/52 – 1/122)

∴ 1/λ = 1.0974 × 107(1/25 – 1/144)

∴ 1/λ = 1.0974 × 107(0.04 – 0.006)

∴ 1/λ = 1.0974 × 107 × 0.034

∴ 1/λ = 0.0373 × 107

∴ λ = 26.8096 × 10-7 m

Question 6: If electrons move from n = 8 to n = 2 then Calculate the wavelength of the photon.

Solution:

Given: n1 = 2, n2 = 8, RH = 1.0974 × 107

Since,

1/λ = RH(1/n12 – 1/n22)

∴ 1/λ = 1.0974 × 107(1/22 – 1/82)

∴ 1/λ = 1.0974 × 107(1/4 – 1/64)

∴ 1/λ = 1.0974 × 107(0.25 – 0.01)

∴ 1/λ = 1.0974 × 107 × 0.24

∴ 1/λ = 0.2633 × 107

∴ λ = 3.7979 × 10-7 m


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