Chapter 27, Problem 39E

(a)

To determine

The energy of lowest level of muonic atom.

Answer to Problem 39E

The energy of lowest energy level is $-7.8\times {10}^5\text{eV}$.

Explanation of Solution

Write the expression for energy of atom.

    $E=-\frac{m_{\mu }{Z}^2{e}^4}{32{\hslash }^2{\pi }^2{\epsilon }_0^2}\frac{1}{n^2}$        (I)

Here, $E$ is the ground state energy, $m_{\mu }$ is the mass of muon, $Z$ is the atomic number, $e$ is the electronic charge, $\hslash$ is reduced Planck’s constant, ${\epsilon }_0$ is the permittivity and $n$ is the number of orbital.

 Conclusion:

Substitute $207\left(9.1\times {10}^{-31}\text{kg}\right)$ for $m_{\mu }$, $1$ for $Z$, $1.6\times {10}^{-19}\text{C}$ for $e$, $1.055\times {10}^{-34}\text{Js}$ for $\hslash$, $8.85\times {10}^{-12}\text{C}/\text{m}$ for ${\epsilon }_0$ and $1$ for $n$ in equation (I).

    $\begin{array}{c}E=-\frac{207\left(9.1\times {10}^{-31}\text{kg}\right){\left(16\right)}^2{\left(1.6\times {10}^{-19}\text{C}\right)}^4}{32{\left(1.055\times {10}^{-34}\right)}^2{\left(3.14\right)}^2{\left(8.85\times {10}^{-12}\text{F}/\text{m}\right)}^2}\frac{1}{{\left(1\right)}^2}\\ =-\frac{207\left(9.1\times {10}^{-31}\text{kg}\right){\left(16\right)}^2{\left(1.6\times {10}^{-19}\text{C}\right)}^4}{32{\left(1.055\times {10}^{-34}\right)}^2{\left(3.14\right)}^2{\left(8.85\times {10}^{-12}\text{F}/\text{m}\right)}^2}\frac{1}{{\left(1\right)}^2}{\left(\frac{1\text{F}}{1\text{C}/\text{V}}\right)}^2\left(\frac{1\text{J}}{1{\text{kgm}}^2{\text{s}}^{-2}}\right){\left(\frac{1\text{J}}{1\text{CV}}\right)}^2\\ =-114.9\times {10}^{-15}\text{J}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =-7.18\times {10}^5\text{eV}\end{array}$ 

Thus, the energy of lowest energy level is $-7.8\times {10}^5\text{eV}$.

(b)

To determine

The radius of lowest energy orbit.

Answer to Problem 39E

The radius of lowest orbit is $1.6\times {10}^{-14}\text{m}$.

Explanation of Solution

Write the expression for radius.

    $r_n=\frac{n^2{\hslash }^2}{kZ{m}_{\mu }{e}^2}$        (II)

Here, $r_n$ is the radius of orbit, $n$ is the number of orbit, $\hslash$ is the reduced Planck’s constant, $k$ is force constant, $m_{\mu }$ is mass of muon, $e$ is electronic charge and $Z$ is the atomic number.

Conclusion:

Substitute $1$ for $n$, $1.055\times {10}^{-34}\text{Js}$ for $\hslash$, $9\times {10}^9\text{F}/\text{m}$ for $k$, $16$ for $Z$, $207\left(9.1\times {10}^{-31}\text{kg}\right)$ for $m_{\mu }$ and $1.6\times {10}^{-19}\text{C}$ for $e$ in equation (II).

    $\begin{array}{c}{r}_1=\frac{{\left(1\right)}^2{\left(1.055\times {10}^{-34}\text{Js}\right)}^2}{\left(9\times {10}^9\text{F}/\text{m}\right)\left(16\right)\left(207\right)\left(9.1\times {10}^{-31}\text{kg}\right){\left(1.6\times {10}^{-19}\text{C}\right)}^2}\\ =\frac{{\left(1\right)}^2{\left(1.055\times {10}^{-34}\text{Js}\right)}^2}{\left(9\times {10}^9\text{F}/\text{m}\right)\left(16\right)\left(207\right)\left(9.1\times {10}^{-31}\text{kg}\right){\left(1.6\times {10}^{-19}\text{C}\right)}^2}\left(\frac{1{\text{kgm}}^2{\text{s}}^{-2}}{1\text{J}}\right)\left(\frac{1\text{F}}{1\text{C}/\text{V}}\right)\left(\frac{1\text{C}/\text{V}}{1\text{J}}\right)\\ =1.6\times {10}^{-14}\text{m}\end{array}$

Thus, the radius of lowest orbit is $1.6\times {10}^{-14}\text{m}$.

(c)

To determine

The ratio of orbital radius to that of sulfur.

Answer to Problem 39E

The ratio of orbital radius to that of sulfur is $4$.

Explanation of Solution

Write the expression for ratio.

    $r=\frac{r_o}{r_s}$        (III)

Here, $r$ is the ratio of radius, $r_o$ is the radius of orbit and $r_s$ is the radius of sulfur.

Conclusion:

Substitute $1.6\times {10}^{-14}\text{m}$ for $r_o$ and $4\times {10}^{-15}\text{m}$ for $r_s$ in equation (III).

    $\begin{array}{c}r=\frac{1.6\times {10}^{-14}\text{m}}{4\times {10}^{-15}\text{m}}\\ =4\end{array}$

Thus, the ratio of orbital radius to that of sulfur is $4$.