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## GR0877 Problem 50

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (C) Dimensional analysis works fine here. However, you can also derive needed expression using Newton’s second law and Bohr’s quantization rule. Indeed, the former gives $mv^2/r = Ze^2/(4 \pi \epsilon_0 r^2)$, while the later $mvr = n \hbar$. From these two equations (excluding $v$): $\displaystyle r_n \propto \frac{n^2}{me^2Z}$. Potential energy: $E_{pot} = -Ze^2/(4 \pi \epsilon_0 r)$. The total energy: $E = E_{kin} + E_{pot} = -Ze^2/(8\pi \epsilon_0 r)$. Substituting $r_n$ into the expression for the total energy yields $\displaystyle E \propto \frac{mZ^2 e^4}{n^2}$. To account for the motion of the nuclei we can treat $m$ in the last equation as the reduced mass. Thus, choice (C).

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1. 14.09.2012 at 04:30

I’m not seeing the dimensional analysis. Z and n are unit-less, so that would make it kg*C^4.

2. 15.09.2012 at 22:06

Well, try another exponents for mass and coulombs, rather than 1 and 4, to get the right energy units (with constants $\hbar$ and $1/4\pi \epsilon_0$ involved)! =) I’m not telling you this is the best solution, but still it can be done.

3. 17.10.2013 at 23:03

I think by dimensional analysis, he doesn’t mean it quite like that, since the problem says “proportional to” so the approach of looking at units isn’t very helpful. Rather, I solved the problem using three facts that you simply have to drill into your head:
1. 1/n^2 should always be expected for energy levels (which you should remember from studying the Hydrogen atom).
2. Mass should only be to the first power (also from the Hydrogen atom).
3. This is the most obscure one to remember, but Z^2 should always appear in energy levels when you’re dealing with any atom bigger than the Hydrogen atom (though it applies to Hydrogen atom too, with Z = 1). I found this one frustrating to remember at first, but it does show up several times throughout the practice tests, so you really just need to memorize it. Or you can think of it like this: there are Z nucleons (that’s the definition of Z) interacting with Z electrons (assuming none are missing or added). If each nucleon interacting with each electron contributes to the total energy (which is true and only ignores electron-electron interactions), then you have a total of Z^2 interactions contributing to the total energy. So to generalize for any value of Z, the energy levels should have a Z^2 dependence.

So back to the problem, you need m*Z^2/n^2, and the only one that fits is C.

• 29.08.2016 at 06:50

For your point #3, you are incorrect. The Bohr-atom is the hydrogen-like atom, which consists of Z nucleons and only one electron (not Z electrons). So your explanation of why there is a Z^2 dependence is wrong.