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Chemistry · Ch 2 — Structure of Atom

Explanation of Line Spectrum of Hydrogen

2.4.1

Explanation of Line Spectrum of Hydrogen

The Line Spectrum of Hydrogen: A Quantitative Explanation

Bohr's model provides a complete quantitative explanation for the line spectrum of hydrogen. The key is the relationship between electron transitions between orbits and the absorption or emission of radiation.

When an electron jumps from a lower energy orbit (smaller principal quantum number nn) to a higher energy orbit (larger nn), the atom absorbs energy. Conversely, when an electron falls from a higher orbit to a lower orbit, the atom emits energy. The energy change involved in any such transition is the difference between the energies of the two orbits involved.

The Energy Gap Between Orbits

The energy of an electron in the nnth orbit of hydrogen is given by:

En=−RHn2E_n = -\frac{R_H}{n^2}

where RH=2.18×10−18 JR_H = 2.18 \times 10^{-18} \text{ J} is the Rydberg constant for hydrogen.

For a transition from an initial orbit nin_i to a final orbit nfn_f, the energy change ΔE\Delta E is:

ΔE=Ef−Ei\Delta E = E_f - E_i

Substituting the energy expression:

ΔE=(−RHnf2)−(−RHni2)\Delta E = \left(-\frac{R_H}{n_f^2}\right) - \left(-\frac{R_H}{n_i^2}\right)

ΔE=RH(1ni2−1nf2)\Delta E = R_H \left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)

This is equation (2.17) in the textbook. In terms of the numerical value of RHR_H:

ΔE=2.18×10−18(1ni2−1nf2) J\Delta E = 2.18 \times 10^{-18} \left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right) \text{ J}

Important

The sign of ΔE\Delta E tells you whether energy is absorbed or emitted:

  • Absorption spectrum: nf>nin_f > n_i, so 1ni2>1nf2\frac{1}{n_i^2} > \frac{1}{n_f^2}, making ΔE\Delta E positive — energy is absorbed.
  • Emission spectrum: ni>nfn_i > n_f, so 1ni2<1nf2\frac{1}{n_i^2} < \frac{1}{n_f^2}, making ΔE\Delta E negative — energy is released.

From Energy to Frequency and Wavenumber

The frequency ν\nu of the photon absorbed or emitted is obtained from Planck's relation E=hνE = h\nu:

ν=∣ΔE∣h=2.18×10−186.626×10−34(1nf2−1ni2) Hz\nu = \frac{|\Delta E|}{h} = \frac{2.18 \times 10^{-18}}{6.626 \times 10^{-34}} \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \text{ Hz}

Evaluating the constant:

ν=3.29×1015(1nf2−1ni2) Hz\nu = 3.29 \times 10^{15} \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \text{ Hz}

This is equation (2.19). Note that for emission, ni>nfn_i > n_f, so the term in parentheses is positive and gives the magnitude of the frequency.

The wavenumber νˉ\bar{\nu} (the reciprocal of wavelength, measured in m−1^{-1}) is related to frequency by νˉ=ν/c\bar{\nu} = \nu/c, where c=3.0×108 m s−1c = 3.0 \times 10^8 \text{ m s}^{-1}:

νˉ=3.29×10153.0×108(1nf2−1ni2) m−1\bar{\nu} = \frac{3.29 \times 10^{15}}{3.0 \times 10^8} \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \text{ m}^{-1}

νˉ=1.09677×107(1nf2−1ni2) m−1\bar{\nu} = 1.09677 \times 10^7 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \text{ m}^{-1}

This is equation (2.21). The constant 1.09677×107 m−11.09677 \times 10^7 \text{ m}^{-1} is the Rydberg constant expressed in wavenumber units.

Note

The expression ΔE=RH(1ni2−1nf2)\Delta E = R_H \left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right) is identical in form to the empirical Rydberg formula (equation 2.9) that was derived from experimental data long before Bohr's model. Bohr's great achievement was to derive this formula theoretically from his postulates, providing a physical explanation for what was previously just an empirical observation. …