Muon ($$\mu^{-1}$$) is negatively charged (|q| = |e|) with a mass m$$_\mu$$ = 200 m$$_e$$, where m$$_e$$ is the mass of the electron and e is the electronic charge. If $$\mu^{-1}$$ is bound to a proton to form a hydrogen like atom, identify the correct statements:
(A) Radius of the muonic orbit is 200 times smaller than that of the electron
(B) The speed of the $$\mu^{-1}$$ in the nth orbit is $$\frac{1}{200}$$ times that of the electron in the nth orbit
(C) The ionization energy of muonic atom is 200 times more than that of an hydrogen atom
(D) The momentum of the muon in the nth orbit is 200 times more than that of the electron

Let us compare an ordinary hydrogen atom (electron + proton) with the “muonic-hydrogen” atom (muon $$\mu^{-}$$ + proton) by using the usual Bohr formulas. Everywhere the electronic charge magnitude is denoted by $$e$$ and the Coulomb constant by $$k=\dfrac{1}{4\pi\varepsilon_{0}}$$.

Bohr first‐postulate formula for the radius of the nth orbit is

$$r_{n}=\frac{n^{2}\hbar^{2}}{k\,m\,e^{2}},$$

where $$m$$ is the mass of the orbiting particle. For an electron we write $$m=m_{e}$$, while for the muon we have $$m=m_{\mu}=200\,m_{e}$$. Taking the ratio of the two radii for the same quantum number $$n$$ we obtain

$$\frac{r_{n}(\text{muon})}{r_{n}(\text{electron})}=\frac{n^{2}\hbar^{2}/(k\,m_{\mu}e^{2})}{n^{2}\hbar^{2}/(k\,m_{e}e^{2})} =\frac{m_{e}}{m_{\mu}} =\frac{1}{200}.$$

So the muonic orbit is 200 times smaller than the electronic one. Statement (A) is therefore correct.

The Bohr expression for the speed in the nth orbit reads

$$v_{n}=\frac{k\,e^{2}}{\hbar}\;\frac{1}{n},$$

which is obtained from the angular-momentum quantisation $$m\,v_{n}\,r_{n}=n\hbar$$ together with the Coulomb force centripetal balance $$\dfrac{m\,v_{n}^{2}}{r_{n}}=\dfrac{k\,e^{2}}{r_{n}^{2}}$$. Notice that in the final form the mass has cancelled out; hence the speed depends only on $$n$$ and is the same for electron and muon:

$$v_{n}(\text{muon})=v_{n}(\text{electron}).$$

Consequently the claim that the muon speed is $$\dfrac{1}{200}$$ of the electron speed is false. Statement (B) is incorrect.

The total energy of the nth orbit in the Bohr model is

$$E_{n}=-\frac{m\,k^{2}e^{4}}{2\hbar^{2}}\;\frac{1}{n^{2}}.$$

The ionisation energy is the magnitude of $$E_{1}$$, so for the ground state we have

$$|E_{1}(\text{muon})|=\frac{m_{\mu}}{m_{e}}\;|E_{1}(\text{electron})| =200\;|E_{1}(\text{electron})|.$$

Thus the muonic atom is 200 times harder to ionise. Statement (C) is correct.

The momentum in the nth Bohr orbit is simply

$$p_{n}=m\,v_{n}.$$

Since we have just seen that $$v_{n}(\text{muon})=v_{n}(\text{electron})$$, substituting $$m_{\mu}=200\,m_{e}$$ gives

$$p_{n}(\text{muon})=m_{\mu}v_{n}=200\,m_{e}\,v_{n}=200\,p_{n}(\text{electron}).$$

Therefore the muon’s orbital momentum is 200 times larger. Statement (D) is also correct.

To summarise, statements (A), (C) and (D) are correct, while (B) is wrong. Among the given choices, this set corresponds to Option 4.

Hence, the correct answer is Option 4.