- #1
squelch
Gold Member
- 57
- 1
Homework Statement
A muon is a particle with a charge equal to that of an electron and a mass equal to 207 times the mass of an electron. What is the radius and energy of the ground state of hydrogen if the electron is replaced with a muon? Don't forget to use the concept of reduced mass.
Homework Equations
[tex]\begin{array}{l}
r = \frac{{4\pi {\varepsilon _0}{\hbar ^2}{n^2}}}{{m{e^2}}}\\
{E_n} = - \frac{{m{e^4}}}{{2{{(4\pi {\varepsilon _0})}^2}{\hbar ^2}{n^2}}}\\
\mu = \frac{1}{{\frac{1}{{{m_1}}} + \frac{1}{{{m_2}}}}} = \frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}
\end{array}[/tex]
The Attempt at a Solution
For the ground state, n=1. To find this hypothetical radius, it seems we simply substitute the mass of the muon in for the mass of the electron, which yields r = 2.53528x10^-13 meters, or 0.0048 the Bohr radius. Similarly for the energy.
However, I'm not precisely sure what to do with the reduced mass. The reduced mass seems to be a product if the mass of both a proton and electron -- is it more appropriate to substitute this reduced mass in for the mass of the electron given the parameters of the problem?