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I read in my textbook (NCERT Textbook in Physics class 12, part 2, chapter 12, p. 432) that some emission lines are stronger than others, due to the preference of some transitions in an atom to others:

While the Bohr’s model correctly predicts the frequencies of the light emitted by hydrogenic atoms, the model is unable to explain the relative intensities of the frequencies in the spectrum. In emission pectrum of hydrogen, some of the visible frequencies have weak intensity, others strong. Why? Experimental observations depict that some transitions are more favoured than others. Bohr’s model is unable to account for the intensity variations.

However, no explanation is provided. How can this be explained?

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  • $\begingroup$ This will be easier to answer if you specify the textbook and passage, and identify the relevant transitions. $\endgroup$ – Emilio Pisanty Mar 10 '17 at 15:17
  • $\begingroup$ It is the NCERT Textbook in Physics, published by the NCERT, India, for use in schools following the CBSE Syllabus - NCERT Physics Class 12 Part 2 Page 432 Paragraph 2. $\endgroup$ – Aaron John Sabu Mar 10 '17 at 15:19
  • $\begingroup$ I am really sorry I can't provide a photograph, but it is an authentic source as it has compiled for school students' use by scientists and professors. $\endgroup$ – Aaron John Sabu Mar 10 '17 at 15:20
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The transitions that arise in the Bohr model of the atom appear experimentally with stronger or weaker signal because the dynamics of the transition are different. To explain this in full you need a proper quantum mechanical calculation, but the bottom line is that transitions have something called an oscillator strength that determines how strongly the transition couples to the electromagnetic field and therefore how strongly the atom will emit in that line.

More generally, though, not all transitions are created equal, but Bohr's model is completely blind to these distinctions; hence the need for a theory that can study those differences.

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To lowest order, the emission rate of photons with polarization ${\boldsymbol \epsilon} $ is proportional to the square of the matrix element $$\mathcal M_{fi}^{(1)} = \langle f | \boldsymbol \epsilon \cdot \mathbf r|i\rangle$$ which, can be expressed as $$\mathcal M_{fi}^{(1)} = \int d^3 r \, \psi^*_f(\mathbf r) \boldsymbol \epsilon \cdot \mathbf r \psi_i(\mathbf r) $$ where $\psi_f, \psi_f$ are the wavefunctions of the final and initial states, respectively. Depending on the particular shapes of the wavefunctions, $\mathcal M_{fi}^{(1)} $ can be larger or smaller. In particular, $\mathcal M_{fi}^{(1)} $ is identically zero unless $\ell_i = \ell_f \pm 1$ where $\ell_{i,f}$ is the initial (final) orbital angular momentum quantum number. This is called a selection rule.

If the selection rule is not fulfilled, the next order contribution to the transition rate is $$\mathcal M_{fi}^{(2)} = i \langle f | (\boldsymbol \epsilon \cdot \mathbf p) (\mathbf k\cdot \mathbf r) | i\rangle$$ where $\mathbf k$ is the wavevector of the transition. Since the wavelengths of atomic transitions are much longer than the size of the atoms (hundreds of nanometers compared to tenths of nanometers), $|\mathbf k \cdot \mathbf r| \ll 1$ wherever the wavefunctions are appreciable, which means that these matrix elements indeed are much smaller than the first order ones. Hence, these transitions are strongly suppressed (but still occur).

In addition, the emission rate contains factors of the energy difference between the initial and final states.

Finally, one must take into account the degeneracies of the states. For example, in the hydrogen atom there are (including spin) 2 $1s$ states and 6 $2p$ states. According to the fundamental assumption of thermodynamics, all $2p$ states are equally populated. Hence, the observed transition rate will be $6/2 = 3$ times higher than if the degeneracy were not taken into account.

A detailed description of how to calculate these things can be found in any introductory quantum mechanics textbook, for example Griffiths or Townsend.

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    $\begingroup$ This may be a bit much for the OP. Have you looked at the text he quotes? $\endgroup$ – Emilio Pisanty Mar 10 '17 at 22:04

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