Physics Asked on December 10, 2021
I’ve been dealing with a couple of questions regarding Bohr’s model except where the mass of the nucleus is not negligible compared to the mass of the (revolving) electron. According to my book (Physics for JEE Advanced by B.M. Sharma), you’re supposed to derive the regular formulae for radius of the nth orbit and energy of an electron in the nth orbit using Bohr’s postulates, and then you’re supposed to replace every $m_{e}$ (mass of the electron) by $mu$, where $mu=frac{m_{n}cdot m_{e}}{m_{n}+m_{e}}$.
I understand why this is true for the total energy of electron in the nth orbit: because multiplying the electron’s kinetic energy in a frame attached to the nucleus by $frac{mu}{m_{e}}$ helps us get the total kinetic energy of the system relative to the center of mass of the system (that $K_{cm}=frac{1}{2}mu v_{rel}^2$ can be derived).
However, for the radius expression, here’s what I tried:
(Let $r_{e}$ be the radius of the electron’s orbit, $r_{n}$ be that of the nucleus, $v_{e}$ be the velocity of the electron and $v_{n}$ that of the nucleus)
$$m_{e}r_{e}=m_{n}r_{n}=lambda$$
$$r_{e}+r_{n}=r$$
$$implies frac{lambda}{m_{e}}+frac{lambda}{m_{n}}=r$$
$$implies frac{lambda}{mu}=r$$
$$implies m_{e}r_{e}=m_{n}r_{n}={mu}r$$
Applying Bohr’s postulate regarding angular momentum:
$$m_{e}(v_{e}+v_{n})r= frac{nh}{2pi} …*$$
$$implies frac{m_{e}^2v_{e}r}{mu}=frac{nh}{2pi}$$
Finally equating centripetal force to electrostatic force:
$$frac{kZe^2}{r^2}=frac{m_{e}v_{e}^2}{r_{e}}$$
$$implies frac{kZe^2}{r^2}=frac{m_{e}^2v_{e}^2}{{mu}r}$$
$$implies {mu kZe^2}=m_{e}^2v_{e}^2r$$
$$implies v_{e} = frac{2pi kZe^2}{nh}$$
Now, using $v_{e}$ to calculate $r$, I got:
$$r = frac{n^2h^2mu}{4pi^2 m_{e}^2 kZe^2}$$
I’m sorry for the long calculations; anyway, according to the book, $r$ should be $$frac{n^2h^2}{4pi^2 mu kZe^2}$$.
Whatever I’ve done seems right to me, but either way, I’ve marked out the line that I’m particularly doubtful of with a $*$ (regarding my usage of Bohr’s third postulate).
TL; DR: How does one calculate the radius of an electron’s orbit (its distance from the nucleus) in terms of the principle quantum number ($n$) if the mass of a nucleus is not negligible enough to consider it at rest with respect to the center of mass?
Your expression for the angular momentum of the system is wrong. It assumes that both the nucleus and the electron are traveling in a a circle of radius $r$; but in reality, the electron travels in a circle of radius $r_e$ and the nucleus travels in a circle of radius $r_n$.
Answered by Michael Seifert on December 10, 2021
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