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I derived the Rydberg constant with the help of Bohr model but I read on the internet that we have to assume the mass of nucleus to be infinitely larger than the electron's. However, I got the answer without any assumption and stuff.

My question is, what is the point of making that assumption. I just used Coulomb's law and quantization of angular momentum.

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  • $\begingroup$ Coulomb's law applied to a circular orbit (fitting of wavelengths) implies this assumption. So you did use it, but you didn't recognize it. $\endgroup$ – John Alexiou Oct 4 '17 at 12:57
  • $\begingroup$ Fitting of wavelengths? I didn't get that $\endgroup$ – Utkarsh Mishra Oct 4 '17 at 18:16
  • $\begingroup$ The orbiting electron has a De-Broglie wavelength $\lambda = {\rm h} / p$ and an integer number of wavelengths have to fit in the orbit. This is the quantization of momentum simplified. $\endgroup$ – John Alexiou Oct 4 '17 at 20:08
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The finite nuclear mass causes a slight variation in the Rydberg constant from atom to atom. In case of finite nuclear mass we have to replace mass of electron by reduced mass of nucleus-electron system. Thus its value is reduced by a factor 1/1+m/M where m is mass of electron and M is mass of nucleus and is different for different atoms,depending upon their nuclear masses. The largest change occurs between H1 and H 2. With increasing mass number,the Rydberg constant approaches more closely to its value for infinitely massive nucleus.

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    $\begingroup$ In case of infinitely massive nucleus you neglect the motion of nucleus and it becomed a one body problem but in case of finite nuclear mass we must take into account the motion of nucleus and so it becomes a two body problem and so reduced mass of the system is taken instead of electronic mass. $\endgroup$ – Rajendra Pd Oct 4 '17 at 7:29

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