91影视

The equilibrium separation distance between the deuterons but$m_e鈫�m_r$the reduced mass of themuon:

role="math" localid="1658383790270" $$\begin{gathered} m_r=\frac{m_{渭}{m}_d}{m_{渭}+m_d} \\ =\frac{m_{渭}2m_p}{m_{渭}+2m_p} \\ =\frac{m_{渭}}{1+m_{渭}/2m_p}. \\ {m}_{渭}=207m_e, \end{gathered}$$

so

$$\begin{gathered} 1+\frac{m_{渭}}{m_p}=1+\frac{207}{2}\frac{\left(9.11脳{10}^{-31}\right)}{\left(1.67脳{10}^{-27}\right)} \\ =1.056;m_r \\ =\frac{207m_e}{1.056}=196m_e \end{gathered}$$

This shrinks the muonic \Bohr radius鈥� down by a factor of nearly 200. but nowis the ground state energy of the muonic atom. The potential energy associated with the deuteron-deuteron repulsion is the same as, and since the productis independent of mass, it doesn鈥檛 matter whether we write it using the electron values or the muon values. the entire molecule shrinks by that same factor of 196. The equilibrium separation for the electron case was 2.493 a (Problem 7.10),Therefore, The general equation of

$$\begin{gathered} R=\frac{2.493}{196}\left(0.529脳{10}^{-10}m\right) \\ =6.73脳{10}^{-10}m \end{gathered}$$

Thus the equilibrium separation distance between deuterons is $R=6.73脳{10}^{-10}m.$