Class 12 > Unit # 26: Atomic Physics > Wavelength of Photon (Rydberg's Formula)

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Wavelength of Photon & Hydrogen Atomic Spectrum

Video Lecture: Spectrum of Hydrogen Atom

Watch the complete lecture explaining the emission of photons during atomic transitions and spectral series:

Watch directly on YouTube (https://youtu.be/4wbxIIblZOc)

Derivation of Wavelength of Photon in Hydrogen Atom

Let us consider a hydrogen atom in which an electron makes a transition from a higher orbit $n$ to a lower orbit $p$. As a result, a photon of frequency $\nu$ and wavelength $\lambda$ is emitted.

According to Bohr's atomic model, the energy of the emitted photon is equal to the energy difference between the two orbits:

$\Delta E = E_n - E_p$

Substituting the expression for the energy of the $n^{\text{th}}$ orbit ($E_n = -\frac{m e^4}{8 \varepsilon_0^2 h^2 n^2}$):

$\Delta E = \left( -\frac{m e^4}{8 \varepsilon_0^2 h^2 n^2} \right) - \left( -\frac{m e^4}{8 \varepsilon_0^2 h^2 p^2} \right)$

Simplifying the signs (since minus minus becomes plus):

$\Delta E = \frac{m e^4}{8 \varepsilon_0^2 h^2 p^2} - \frac{m e^4}{8 \varepsilon_0^2 h^2 n^2}$

Taking common terms out:

$\Delta E = \frac{m e^4}{8 \varepsilon_0^2 h^2} \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$

According to Planck's quantum theory, the energy of a photon is given by $\Delta E = h\nu$. Substituting this:

$h\nu = \frac{m e^4}{8 \varepsilon_0^2 h^2} \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$

$\nu = \frac{m e^4}{8 \varepsilon_0^2 h^3} \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$

As we know that the relation between frequency ($\nu$), speed of light ($c$), and wavelength ($\lambda$) is $\nu = \frac{c}{\lambda}$, substituting this into the equation:

$\frac{c}{\lambda} = \frac{m e^4}{8 \varepsilon_0^2 h^3} \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$

$\frac{1}{\lambda} = \frac{m e^4}{8 \varepsilon_0^2 h^3 c} \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$

Where the constant term is defined as Rydberg's Constant ($R_H$):

$R_H = \frac{m e^4}{8 \varepsilon_0^2 h^3 c} \approx 1.097 \times 10^7\text{ m}^{-1}$

Therefore, we arrive at the final wave number equation:

$\frac{1}{\lambda} = R_H \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$

Hydrogen Spectral Series

Depending on the lower energy level ($p$) to which the electron transitions, different spectral series are observed:

• Lyman Series ($p = 1$): Ultraviolet region (transitions to the 1st orbit).
• Balmer Series ($p = 2$): Visible region (transitions to the 2nd orbit).
• Paschen Series ($p = 3$): Infrared region (transitions to the 3rd orbit).
• Brackett Series ($p = 4$): Infrared region (transitions to the 4th orbit).
• Pfund Series ($p = 5$): Infrared region (transitions to the 5th orbit).

Fig: Electron transitions from higher states to lower orbits forming spectral series.

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