Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom i.e. an atom where the electron is replaced by a negatively charged muon (\mu ^{-})  of mass about 207\; m_{e} that orbits around a proton. (Given for hydrogen atom, radius of first orbit and ground state energy are 0.53\times 10^{-10}m and -13.6\; eV respectively)

 

 
 
 
 
 

Answers (1)
S safeer

To determine the radius of I^{st} orbit of the muonic hydrogen atom,

We know, In Bohr's model of the hydrogen atom, the radius of n^{th} orbit is given as ;

 r_{n}=\frac{\epsilon_0n^{2}h^{2}}{\pi me^{2}}

 h= plank's constant

 e= charge of electron

  n= no. of orbit

as; n=1  then, r_{1}\propto \frac{1}{m}

Similarly for the muonic hydrogen atom.

        r_{\mu }\propto \frac{1}{m_{\mu }}

Hence,

        \frac{r_{\mu }}{r_{e}}=\frac{m_{e}}{m_{\mu }}=\frac{1}{207}\; \; \; \; \left ( \because m_{u}=207\right )

    \therefore r_{\mu }=2.56\times 10^{-13}m

Now, the energy of the electron in n^{th} orbit ;

E_{n}=\frac{-{me}^{4}}{8\epsilon_0^2 n^{2}h^{2}}

\because \; E_{n}\propto m_{e}             Therefore, 

\therefore \; \frac{E_{\mu }}{E_{e}}=\frac{m_{\mu }}{m_{e}}=207

Where, m_{\mu }= mass of muonic

              m_{e }= mass of electron

\therefore \; E_{\mu }=207\; E_{e}

              =-207\times 13.6\; eV

      E_{\mu }=2.8\; KeV, is the ground state energy of a muonic hydrogen atom.

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