Chapter 27, Problem 37E

(a)

To determine

The ground state energy of positronium.

Answer to Problem 37E

The ground state energy of positronium is $-6.8\text{eV}$.

Explanation of Solution

Write the expression for ground state of energy.

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

Here, $E$ is the ground state energy, $\mu$ is the reduced mass, $Z$ 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.

Positronium consists of two electrons.

Write the expression for reduced mass of positronium.

    $\begin{array}{c}\mu =\frac{m_e{m}_e}{m_e+m_e}\\ =\frac{m_e^2}{2m_e}\end{array}$

    $\mu =\frac{m_e}{2}$        (II)

Here, $\mu$ is the reduced mass and $m_e$ is the mass of electron.

Conclusion:

Substitute $9.1\times {10}^{-31}\text{kg}$ for $m_e$ in equation (II).

    $\begin{array}{c}\mu =\frac{9.1\times {10}^{-31}\text{kg}}{2}\\ =4.55\times {10}^{-31}\text{kg}\end{array}$

Substitute $4.55\times {10}^{-31}\text{kg}$ for $\mu$, $1$ for $Z$, $1.6\times {10}^{-19}\text{C}$ for $e$, $1.055\times {10}^{-34}\text{Js}$ for $\hslash$, $8.85\times {10}^{-12}\text{F}/\text{m}$ for ${\epsilon }_0$ and $1$ for $n$ in equation (I).

    $\begin{array}{c}E=-\frac{\left(4.55\times {10}^{-31}\text{kg}\right){\left(1\right)}^2{\left(1.6\times {10}^{-19}\text{C}\right)}^4}{32{\left(1.055\times {10}^{-34}\text{Js}\right)}^2{\left(3.14\right)}^2{\left(8.85\times {10}^{-12}\text{F}/\text{m}\right)}^2}\frac{1}{{\left(1\right)}^2}\\ =-\frac{\left(4.55\times {10}^{-31}\text{kg}\right){\left(1\right)}^2{\left(1.6\times {10}^{-19}\text{C}\right)}^4}{32{\left(1.055\times {10}^{-34}\text{Js}\right)}^2{\left(3.14\right)}^2{\left(8.85\times {10}^{-12}\text{F}/\text{m}\right)}^2}\frac{1}{{\left(1\right)}^2}{\left(\frac{1\text{F}}{1\text{C}/\text{V}}\right)}^2\left(\frac{1\text{J}}{1{\text{kgm}}^2{\text{s}}^{-2}}\right){\left(\frac{1\text{J}}{1\text{CV}}\right)}^2\\ =-1.08\times {10}^{-18}\text{J}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =-6.8\text{eV}\end{array}$

Thus, the ground state energy of positronium is $-6.8\text{eV}$.    

(b)

To determine

The radius of lowest energy orbit.

Answer to Problem 37E

the radius of lowest orbit of positronium is $1.06\times {10}^{-10}\text{m}$.

Explanation of Solution

Write the expression for the radius of atom.

    $r_n=\frac{{\hslash }^2{n}^2}{\mu k{e}^{2}Z}$        (I)

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

Write the expression for reduced mass of positronium.

    $\begin{array}{c}\mu =\frac{m_e{m}_e}{m_e+m_e}\\ =\frac{m_e^2}{2m_e}\end{array}$

    $\mu =\frac{m_e}{2}$        (II)

Here, $\mu$ is the reduced mass and $m_e$ is the mass of electron.

Conclusion:

Substitute $9.1\times {10}^{-31}\text{kg}$ for $m_e$ in equation (II).

    $\begin{array}{c}\mu =\frac{9.1\times {10}^{-31}\text{kg}}{2}\\ =4.55\times {10}^{-31}\text{kg}\end{array}$

Substitute $1.055\times {10}^{-34}\text{Js}$ for $\hslash$, $1$ for $n$, $4.55\times {10}^{-31}\text{}\text{kg}$ for $\mu$, $9\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^2$ for $k$, $1.6\times {10}^{-19}\text{C}$ for $e$ and $1$ for $Z$ in equation (I).

    $\begin{array}{c}{r}_1=\frac{{\left(1.055\times {10}^{-34}\text{Js}\right)}^2{\left(1\right)}^2}{\left(4.55\times {10}^{-31}\text{kg}\right)\left(9\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^2\right){\left(1.6\times {10}^{-19}\text{C}\right)}^2\left(1\right)}\\ =\frac{{\left(1.055\times {10}^{-34}\text{Js}\right)}^2{\left(1\right)}^2}{\left(4.55\times {10}^{-31}\text{kg}\right)\left(9\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^2\right){\left(1.6\times {10}^{-19}\text{C}\right)}^2\left(1\right)}{\left(\frac{1{\text{kgm}}^2{\text{s}}^{-2}}{1\text{J}}\right)}^2\\ =1.06\times {10}^{-10}\text{m}\end{array}$

Thus, the radius of lowest orbit of positronium is $1.06\times {10}^{-10}\text{m}$.