Question

The radius of the first orbit of hydrogen is ${{r}_{H}}$, and the energy in the ground state is $-13.6eV$. Considering a $\mu -$ particle with a mass $207{{m}_{e}}$ revolving round a proton as in a Hydrogen atom. The energy and radius of proton and $\mu -$ combination respectively in the first orbit are:
[Assume nucleus to be stationary]
(A). $-13.6\times 207eV,\,207{{r}_{H}}$
(B). $-13.6\times 207eV,\dfrac{{{r}_{H}}}{207}$
(C). $207-13.6eV,\,207{{r}_{H}}$
(D). $207-13.6eV,\,\dfrac{{{r}_{H}}}{207}$

Answer 1

Hint: When the electron in a Bohr’s model of hydrogen is replaced by $\mu -$, all parameters which depend on mass change as it has a greater mass than electron. We can calculate the expression for radius and energy by using properties of Bohr’s model and replace the mass of the electron by the mass of $\mu -$. The permissible orbits have electrons with angular momentum $\dfrac{h}{2\pi }$. The electrostatic force of attraction provides the centripetal force between the nucleus and electron.

Formula used:
$mvr=\dfrac{nh}{2\pi }$
$\dfrac{m{{v}^{2}}}{r}=\dfrac{z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$

Complete step-by-step solution:
In a Bohr’s atom, the permissible orbits are those in which the angular momentum of electrons is an integral multiple of $\dfrac{h}{2\pi }$. Therefore,
$mvr=\dfrac{nh}{2\pi }$ - (1)
Here,
$m$ is the mass of electron in the orbit
$v$ is the velocity of electron
$n$ is an integer which depicts the number of orbit
$r$ is the radius of the orbit

The centripetal force is provided by the electrostatic force of attraction between the nucleus and electron. Therefore,
$\dfrac{m{{v}^{2}}}{r}=\dfrac{z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$ - (2)
Here,
$z$ is the atomic number
$e$ is the magnitude of charge
${{\varepsilon }_{0}}$ is permittivity of free space

Using eq (1) and eq (2), the expression for radius in Bohr’s model of atom is-
$r=\left( \dfrac{{{h}^{2}}{{\varepsilon }_{0}}}{\pi mz{{e}^{2}}} \right)\dfrac{{{n}^{2}}}{z}$

Therefore, from the above equation, $r\propto \dfrac{1}{m}$. Therefore, if we replace electron by $\mu -$ the mass is
$m'=207m$. Replacing $m$ with $m'$ in above equation, we get,
$\begin{align}
  & r=\left( \dfrac{{{h}^{2}}{{\varepsilon }_{0}}}{\pi m'z{{e}^{2}}} \right)\dfrac{{{n}^{2}}}{z} \\
 & \Rightarrow r=\left( \dfrac{{{h}^{2}}{{\varepsilon }_{0}}}{\pi 207mz{{e}^{2}}} \right)\dfrac{{{n}^{2}}}{z} \\
 & \therefore r=\left( \dfrac{{{h}^{2}}{{\varepsilon }_{0}}}{\pi mz{{e}^{2}}} \right)\dfrac{{{n}^{2}}}{207z} \\
\end{align}$

The radius for hydrogen is
${{r}_{H}}=\left( \dfrac{{{h}^{2}}{{\varepsilon }_{0}}}{\pi z{{e}^{2}}} \right)\dfrac{{{(1)}^{2}}}{1}$
$\therefore {{r}_{H}}=\left( \dfrac{{{h}^{2}}{{\varepsilon }_{0}}}{\pi z{{e}^{2}}} \right)$ - (3)

The radius for $\mu -$ is-
$\begin{align}
  & r'=\left( \dfrac{{{h}^{2}}{{\varepsilon }_{0}}}{\pi mz{{e}^{2}}} \right)\dfrac{{{(1)}^{2}}}{207(1)} \\
 & \therefore r'=\dfrac{1}{207}\left( \dfrac{{{h}^{2}}{{\varepsilon }_{0}}}{\pi mz{{e}^{2}}} \right) \\
\end{align}$
Therefore, from eq (3),
$r'=\dfrac{{{r}_{H}}}{207}$
The radius for $\mu -$ is $\dfrac{{{r}_{H}}}{207}$.

Using eq (1) and eq (2), the expression for total energy of an electron in its orbit is-
$E=\left( -\dfrac{m{{e}^{4}}}{8\varepsilon _{0}^{2}{{h}^{2}}} \right)\dfrac{1}{{{n}^{2}}}$ - (5)

If we replace the electron by $\mu -$ particle mass changes to $m'=207m$. Replacing $m$ with $m'$ in above equation, we get,
$E'=\left( -\dfrac{207m{{e}^{4}}}{8\varepsilon _{0}^{2}{{h}^{2}}} \right)\dfrac{1}{{{n}^{2}}}$
$\Rightarrow E'=\left( -\dfrac{m{{e}^{4}}}{8\varepsilon _{0}^{2}{{h}^{2}}} \right)\dfrac{207}{{{n}^{2}}}$ - (6)

The energy of electron in hydrogen atom in the first orbit is-
$E=\left( -\dfrac{m{{e}^{4}}}{8\varepsilon _{0}^{2}{{h}^{2}}} \right)\dfrac{1}{{{(1)}^{2}}}=-13.6eV$

Energy of $\mu -$ in the first orbit of hydrogen atom is-
$E'=\left( -\dfrac{m{{e}^{4}}}{8\varepsilon _{0}^{2}{{h}^{2}}} \right)\dfrac{207}{{{(1)}^{2}}}=-13.6\times 207eV$

Therefore, the radius of the $\mu -$ and proton combination for the first orbit of hydrogen is $\dfrac{{{r}_{H}}}{207}$ and the energy is $-13.6\times 207eV$.

Hence, the correct option is (B).

Note:
Bohr’s model of atom is only applicable for those atoms in which only one electron is present. The energy is not released by the electron when revolving in an orbit in Bohr’s model and hence the atom is stable. When an electron moves from one orbit to another, energy possessed by it changes and hence energy is absorbed or released.