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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)

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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Safeer PP

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