Imagine that the electron in a hydrogen atom is replaced by a muon ($$\mu$$). The mass of muon particle is 207 times that of an electron and charge is equal to the charge of an electron. The ionization potential of this hydrogen atom will be:

In the Bohr model, the energy of an electron in the ground state of a hydrogen atom is given by $$E_n = -\frac{m e^4}{8\varepsilon_0^2 h^2 n^2}$$, where $$m$$ is the mass of the orbiting particle, $$e$$ is its charge, $$\varepsilon_0$$ is the permittivity of free space, and $$h$$ is Planck's constant. For the ground state ($$n = 1$$), the ionization energy equals $$|E_1| = \frac{m e^4}{8\varepsilon_0^2 h^2}$$.

This expression shows that the ionization energy is directly proportional to the mass of the orbiting particle (assuming the nucleus is infinitely heavy compared to the orbiting particle, so we do not need the reduced mass correction). For a standard hydrogen atom with an electron of mass $$m_e$$, the ionization energy is $$13.6$$ eV.

When the electron is replaced by a muon with mass $$m_\mu = 207\,m_e$$ and the same charge as the electron, every factor in the energy formula stays the same except $$m$$, which increases by a factor of 207. Therefore the ionization energy of muonic hydrogen is $$E_\mu = 207 \times 13.6 = 2815.2$$ eV.