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While reading about the Bohr model I noticed the following equation: $$ {m_\mathrm{e} v^2\over r} = {Zk q_e^2 \over r^2} $$ Is it possible to explain which formula they used? How from the speed and radius they got an equation depending $Z$ (the number of protons)?

I understand that $F=ma$ and then $\frac{kq_e^2}{r^2}=\frac{mv^2}{r}$ but where $Z$ came from?

EDIT: Is it because $F=\frac{kq_eq_p}{r^2}$ and because $q_p=z\cdot q_e$ we get $F=\frac{kq_e\cdot z\cdot q_e}{r^2}=\frac{kzq_e^2}{r^2}$? My question is if $q_p=z\cdot q_e$ is true.

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  • $\begingroup$ Where did you see the second one? $\endgroup$
    – Urb
    Commented Oct 9, 2020 at 15:06
  • $\begingroup$ @Urb In a cheetsheet someone made preparing for the exam. The second one may be a mistake. $\endgroup$
    – vesii
    Commented Oct 9, 2020 at 15:07
  • $\begingroup$ Don't post a question based on someone's incorrect cheatsheet! Go ask the person that made that mistake. $\endgroup$
    – Bill N
    Commented Oct 9, 2020 at 15:47
  • $\begingroup$ @BillN The cheetsheet exists for some while now so I don't know who is the author. Anyway, I'll remove the second equation, although the question still remains. The question is if $q_p=z\cdot q_e$ it true and why. $\endgroup$
    – vesii
    Commented Oct 9, 2020 at 15:56

1 Answer 1

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OP's second equation (v2) is indeed wrong as it not dimensionally correct. In the first equation, the $Z$ comes from the fact that Bohr's model considers not only hydrogen atoms, where the nucleus has charge $+e$, but also hydrogen-like atoms, that have only one electron but charge $+Ze$ in the nucleus, like ionized helium $\require{mhchem}\ce{He^+}$, doubly ionized lithium $\ce{Li^{++}}$, etc.

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  • $\begingroup$ So as I understand the colon power is $F=\frac{kq_eq_p}{r^2}$ and because $q_p=z\cdot q_e$ we get $F=\frac{kq_e\cdot z\cdot q_e}{r^2}=F=\frac{kzq_e^2}{r^2}$? $\endgroup$
    – vesii
    Commented Oct 9, 2020 at 15:37
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    $\begingroup$ If by "colon" you mean Coulomb (the colon is another thing) and by "power" you mean force, then yes (sort of). That notation is misleading; don't confuse $q_p$ with the charge of a proton. The charge of a proton is $+e$. What you denote by $q_p$ is the charge of the nucleus, which has $Z$ protons. $\endgroup$
    – Urb
    Commented Oct 9, 2020 at 15:58
  • $\begingroup$ Oh that makes much more sense! Thank you! $\endgroup$
    – vesii
    Commented Oct 9, 2020 at 16:03

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