Can anyone confirm the ground state Bohr radius for muonic hydrogen?

[seattle.truth]

Thanks in advance for anybody who is kind enough to help me. No this isn't for my homework. I am not even enrolled in school. I am doing some calculations for personal research.

But I need to know the ground state radius of a muonic hydrogen atom to help prove my theory.

I already know the equation \frac{\hbar^{2}}{m_{m}e^{4}} .

I don't need help with finding the equation. I just need somebody to confirm the number so I know I didn't mess up (possibly from an outside source if you know where to find it). So please just post the length of the radius, I just need the number.

Maybe I am a dumbass. I searched all over the internet and could not find a clear answer.

Sorry if that's the case, but I really appreciate anyone who can take a minute to help me out.

Thanks

 

[fzero]

Thanks in advance for anybody who is kind enough to help me. No this isn't for my homework. I am not even enrolled in school. I am doing some calculations for personal research.

But I need to know the ground state radius of a muonic hydrogen atom to help prove my theory.

I already know the equation \frac{\hbar^{2}}{m_{m}e^{4}} .

I don't need help with finding the equation. I just need somebody to confirm the number so I know I didn't mess up (possibly from an outside source if you know where to find it). So please just post the length of the radius, I just need the number.

Maybe I am a dumbass. I searched all over the internet and could not find a clear answer.

Sorry if that's the case, but I really appreciate anyone who can take a minute to help me out.

Thanks

The Bohr radius is given by

[tex] a_0 = \frac{ 4\pi \epsilon_0\hbar^2 }{\bar{m}_e e^2} = \frac{ \hbar } {\bar{m}_e c \alpha},[/tex]

where [tex]\alpha[/tex] is the fine-structure constant and [tex]\bar{m}_e[/tex] is the reduced mass of the electron:

[tex] \bar{m}_e = \frac{ m_e m_p}{m_e+m_p}.[/tex]

Typically we replace [tex]\bar{m}_e[/tex] with [tex]m_e[/tex] because the error in doing so is very small (0.2% or so), but the muon is quite a bit more massive in relation to the proton, so we must use the reduced mass. For a muon

[tex] a_\mu = \frac{ \hbar } {\bar{m}_\mu c \alpha},[/tex]

or we can write

[tex] \frac{a_\mu}{a_0} = \frac{\bar{m}_e}{\bar{m}_\mu} = \frac{m_e}{m_\mu} \frac{m_\mu + m_p}{m_e+m_p} .[/tex]