# Order of magnitude of $v=\sqrt{q^2/Mr}$

I'm developing a problem and already found the solution, I know is correct to assume in my problem that the velocity of an electron is:

$$v=\sqrt{\frac{q^2}{Mr}},$$

where $$q$$ is the charge of the electron, $$M$$ is it's mass and $$r$$ is a radius of $$0.5$$Å.

I have computed and obtained $$\sim 10^1$$ but I should obtain $$\sim 10^6$$.

I don't understand what is happening and is driving me crazy.

• Expression units doesn't match for speed. So seems there is an error in the formula. How you derived it ? Commented Dec 4, 2023 at 16:15
• You are missing $(4\pi\epsilon_0)^{-1}$ inside the square root.
– Puk
Commented Dec 4, 2023 at 16:59

There's several ways to derive classic electron speed in a hydrogen atom, based on simplified Bohr atom model. For example, you can equate Coulomb force work done rotating electron around nuclei to the electron's kinetic energy (it may not be ideal way of solution, but gives same answer) :

$$F_e \cdot 2 \pi r = \frac {m_e v^2}{2} ,\tag 1$$

The reason of why we can do this is that Coulomb force gives to the electron such kinetic energy for rotating complete cycle around nuclei with some tangential speed. Substituting electrostatic force into $$F_e$$ and solving for speed gives :

$$v = \sqrt{\frac {4 \pi k_e q^2}{m_e r}} \tag 2,$$

So you miss part $$\{4 \pi k_e\}$$ in numerator, where $$k_e$$ is Coulomb constant. From (2) follows that approximate tangential electron speed in $$\text{H}$$ atom is $$\approx 8 \times 10^{6}~m/s$$ or about $$3\%$$ of light speed in vacuum.

• Right, forgot the Coulomb constant. Thank you very much! Commented Dec 4, 2023 at 19:55
• 3% of light speed in vacuum That’s not correct. You should get $v/c=\alpha\approx 1/137$ for the ground state of the Bohr model. Commented Dec 4, 2023 at 20:10
• Thanks, seems $4\pi$ factor messes everything up,- without it answer would be exactly $\alpha \cdot c$. So yes, my approach is not best to find out exact $v/c$ ratio. On the other hand OP is asking about speed order of magnitude. Both cases - your and mine give same order of magnitude,- $10^6~m/s$. Commented Dec 5, 2023 at 8:15
• it may not be ideal way of solution Actually this solution makes no sense (except dimensionally). There is no “Coulomb force work done” on the electron because it is always moving at right angles to the force on it. If you want to just make a dimensional argument, you shouldn’t base it on false and highly misleading physics like “Coulomb force gives to the electron such kinetic energy for rotating complete cycle around nuclei with some tangential speed”. Commented Dec 5, 2023 at 18:04
• Yes, the Coulomb force makes the electron move in a circle (in the incorrect Bohr model). No, it doesn’t do any work on the electron. Please refer to Wikipedia for the definition of work. It should be clear that when a radial force causes circular motion, $\mathbf F\cdot\mathbf v$ is zero since the two vectors are orthogonal, and thus its time integral — which is the work — is zero. I can’t teach classical mechanics in comments, so I’m done. The real lesson here is that what seems “natural” is often incorrect. Commented Dec 5, 2023 at 20:37