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**Hint:**First calculate the reduced mass of the system, that is, the reduced mass of the $\mu - meson$ and proton. With the help of reduced mass, Bohr’s first orbit radius and the binding energy can be calculated for the hydrogen-like atom.

**Complete step by step answer:**

Step 1:

First calculate the reduced mass of the system. Express the formula for the reduced mass of the two-particle system.

$\therefore \mu = \dfrac{{{m_{\text{meson}}} \times {m_p}}}{{{m_{\text{meson}}} + {m_p}}}$ , where ${m_{\text{meson}}}$ is the mass of the $\mu - meson$ and ${m_p}$ is the mass of the proton.

Substitute ${m_{\text{meson}}} = 270m$ and ${m_p} = 1836m$ , where $m$ is the mass of the electron.

$\therefore \mu = \dfrac{{270m \times 1836m}}{{270m + 1836m}}$

$ \Rightarrow \mu = 186m$

Now express the formula for Bohr’s first orbit radius.

$\therefore {r_1} = \dfrac{{{h^2}{1^2}}}{{4{\pi ^2}k{e^2}Z}}$

Now substitute $Z = 186m$ in the above expression.

$\therefore {r_1} = \dfrac{{{h^2}}}{{4{\pi ^2}k{e^2}186m}}$

$ \Rightarrow {r_1} = \dfrac{1}{{186}} \times 0.529\text{angstrom}$

$ \Rightarrow {r_1} = 0.002844\text{angstrom}$

This is the radius of the Bohr’s orbit of the system.

Step 2:

Now express the formula for the binding energy of the hydrogen-like atom's first orbit.

\[\therefore {E_1} = \dfrac{{2\pi {k^2}{e^4}\mu }}{{{h^2}{1^2}}}\]

Substitute the value of $\mu $ .

\[\therefore {E_1} = \dfrac{{2\pi {k^2}{e^4}186m}}{{{h^2}{1^2}}}\]

We know that the binding energy for the first orbit of the hydrogen \[{E_{hydrogen}} = \dfrac{{2\pi {k^2}{e^4}m}}{{{h^2}{1^2}}} = 13.6eV\] , therefore,

$\therefore {E_1} = 186 \times 13.6eV$

${E_1} = 186 \times 13.6eV$

This is the binding energy of the hydrogen-like atom.

Step 3:

In the Layman series, the initial position of the electron may be ${n_i} = 2,3,4,........$ . The transition ends when the electron reaches the ground state. Express the formula for the wavelength of the line in the Lyman series for hydrogen-like atoms.

$\therefore \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{n_f}^2}} - \dfrac{1}{{{n_i}^2}}} \right)$ , where $R$ is Rydberg constant, ${n_f}$ is the final position of the electron, and ${n_i}$ is the initial position of the electron. Substitute Rydberg constant for the reduced mass, ${n_f} = 1$ and ${n_i} = 2$

$\therefore \dfrac{1}{\lambda } = {R_\mu }\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right)$

$ \Rightarrow \dfrac{1}{\lambda } = \dfrac{3}{4}{R_\mu }$

$ \Rightarrow \lambda = \dfrac{4}{{3{R_\mu }}}$ ----(1)

Step 4:

We know that the Rydberg constant for reduced mass is expressed as

$\therefore {R_\mu } = \dfrac{{2{\pi ^2}{k^2}{e^2}\mu }}{{c{h^3}}}$

Substitute the value of $\mu $ in the above expression

$\therefore {R_\mu } = \dfrac{{2{\pi ^2}{k^2}{e^2}186m}}{{c{h^3}}}$

$ \Rightarrow {R_\mu } = \dfrac{{2{\pi ^2}{k^2}{e^2}m}}{{c{h^3}}} \times 186$

$\therefore {R_\mu } = R186$

Substitute the value of ${R_\mu }$ in the equation (1).

$\therefore \lambda = \dfrac{4}{{3 \times 186R}}$

Substitute the value $R = 10967800{m^{ - 1}}$ given in the question.

$\therefore \lambda = \dfrac{4}{{3 \times 186 \times 10967800}}$

$\lambda = 653.6\text{angstrom}$

This is the wavelength of the line in the Lyman series.

**Note:**It is a quantity that allows the two-body problem to be solved as if it were a one-body problem. Note, however, that the mass determining the gravitational force is not reduced. In the computation, one mass can be replaced with the reduced mass, if this is compensated by replacing the other mass with the sum of both masses.

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