Is muonic hydrogen atom or exotic hydrogen atom more stable

According to Bohr's model of hydrogen atom, the energy of the electron is given by: $$E=\large-\frac{me⁴}{8εh²n²}$$ So if we replaced the electron with a muon whose mass is ~200me, will that new atom be more stable than usal hydrogen atom? Because the new atom has even lower potential energy.

• unfortunately for your argument, muons are unstable with a half life measured in $\mu s$
– jim
May 18 at 10:07
• @jim but that half life is for an isolated muon. But in this case it under the interaction with proton. Similarly a nuclei would jave been unstable because the half life of neutron is 1000s. But that again is when neutron is isolated. So does that mean that coulumbic force doesn't impact the half life of muon? Because if the muonic hydrogen atom decays in microseconds then this should be true. May 18 at 10:54
• If muons were stable as part of a hydrogen atom, we would have exploited this fact to achieve room temperature hydrogen fusion already. youtu.be/aDfB3gnxRhc May 18 at 20:59

The attached figure shows the Coulomb potential between a proton and an electron $$V = -\frac{ e^2 }{ 4 \pi \epsilon_0 r }.$$ Substituting for the charges and measuring r in terms of the Bohr radius $$r \approx 0.5 nm$$ gives $$V = -\frac{ 13.6 }{ r } eV,$$ where the Bohr radius is given as $$a_0 = \frac{ 4 \pi \epsilon_0 \hbar^2}{ m_e e^2}$$ with $$m_e$$ is the electron mass.

Note that this system is stable in the sense that when the electron is in its ground state energy $$( \approx -13.6 eV )$$ the electron will never be able to leave the influence of the proton, it is bound and would require someone to give the electron at least $$13.6 eV$$ of energy. You are correct in suggesting that if you replace the electron by a muon, then the muon is more bound than the electron, roughly by the factor $$m_\mu/m_e \approx 200,$$ so that the ground state binding energy will be of order $$200 \times 13.6 eV = 2700 eV.$$

However, I believe that the muon will still decay in a time of order$$\mu s.$$ Let me explain this by looking at the deuteron: The deuteron is the stable bound state that consists of a proton and a neutron. A free neutron is known to be unstable with a half life of order $$13 min$$, so why is the deuteron stable? The reason for this is when a proton and neutron bind some $$2.3 MeV$$ of energy is radiated away, that is the mass of the deuteron is some $$2.3 MeV$$ lighter than the constituent particles.

The free neutron decays into a proton and an electron (other particles are involved) and the mass difference between the neutron and (proton and electron) is about $$0.8 MeV.$$ It is no longer energetically possible for the neutron to decay.

For the case of the muon, the free muon decays into an electron (and other particles not of interest for the moment). But the energy difference is now $$\approx 199 m_e\approx 100 MeV.$$

By comparison, the binding energy of the muon and proton is only measured in thousands of $$eV$$ and it is still energetically feasible for the muon to decay even though bound by the proton.

• Two minor comments. (1) Units should be set in upright type and have a thin space separating them from their numbers; try e.g. $-13.6\ \mathrm{eV}$. (2) Not that it’s a big deal to people who didn’t spend a decade doing $\rm np \to d\gamma$, but the deuteron’s binding energy is closer to 2.2 MeV than to 2.3 MeV.
– rob
May 18 at 21:55
• That quibble aside, this answer is absolutely correct. Muonic hydrogen is more tightly bound in its ground state than ordinary hydrogen, because the muon is heavier. (The $1/m_e$ in the length scale is more properly the “reduced mass” of the two particles.) But the muon-proton binding energy is still a small correction to the muon’s weak decay energy, so the decays of muons in muonic hydrogen are not very different from the decays of free muons.
– rob
May 18 at 22:00