Chapter 4, Problem 56P

(a)

To determine

The energies of the three lowest states of positronium.

Answer to Problem 56P

The energies of the three lowest states of positronium is $-6.803\text{eV}$, $-1.701\text{eV}$ and $-0.756\text{eV}$.

Explanation of Solution

Positronium is an atom composed of an electron and a Positron and the mass of positron is equal to the mass of electron.

Write the expression to calculate reduced mass.

$\mu =\frac{mM}{m+M}$        (I)

Here, $\mu$ is the reduced mass of an atom, $m$ is the mass of positron and $M$ is the mass of electron.

Since the mass of positron is equal to the mass of electron above equation can be written as:

    $\begin{array}{c}\mu =\frac{mm}{m+m}\\ =\frac{m}{2}\end{array}$

Write the expression for effective Rydberg constant.

    $R_{\text{eff}}=\frac{R_{\infty }}{2}$

Here, $R_{\text{eff}}$ is the effective Rydberg constant and $R_{\infty }$ is the Rydberg constant for infinite nuclear mass.

Write the expression for energy at $n^{th}$ level.

    $E_n=\frac{-hc\left(R_{\text{eff}}\right)}{n^2}$        (II)

Here, $E_n$ is the energy at $n^{th}$ level, $h$ is the Planck’s constant, and $c$ is the speed of light.

Conclusion:

Substitute $1239.8\text{eV}\cdot \text{nm}$ for $hc$ , $1.097373\times {10}^{-2}/2{\text{nm}}^{\text{-1}}$ for $R_{\text{eff}}$ and $1$ for $n$ in equation (II).

    $\begin{array}{c}{E}_1=\frac{-\left(1239.8\text{eV}\cdot \text{nm}\right)\left(\frac{1.097373\times {10}^{-2}}{2}{\text{nm}}^{\text{-1}}\right)}{1^2}\\ =-6.803\text{eV}\end{array}$

Similarly for $n=2$ energy is $E_2=-1.701\text{eV}$, for $n=3$ energy is $E_3=-0.756\text{eV}$ and for $n=4$ energy is $E_4=-0.425\text{eV}$

Thus, the energies of the three lowest states of positronium is $-6.803\text{eV}$, $-1.701\text{eV}$ and $-0.756\text{eV}$.

(b)

To determine

The wavelengths of the $K_{\alpha },K_{\beta },L_{\alpha }\text{and}{\text{L}}_{\beta }$ transitions.

Answer to Problem 56P

The wavelengths of the $K_{\alpha }$ transition is $243.0\text{nm}$ $,K_{\beta }$ transitions is $205.0\text{nm}$ $,L_{\alpha }$ transition is $1312\text{nm}$ and ${\text{L}}_{\beta }$ transition is $971.6\text{nm}$.

Explanation of Solution

Write the expression for wavelength.

    $\lambda =\frac{hc}{E_u-E_l}$        (III)

Here, $\lambda$ is the wavelength of transition, $E_u$ is energy of upper energy state and $E_l$ is the energy of lower energy state.

Conclusion:

For $\lambda \left(K_{\alpha }\right)$ : $n_l=1\text{and}{n}_u=2$.

Substitute $1239.8\text{eV}\cdot \text{nm}$ for $hc$, $-1.701\text{eV}$ for $E_u$ and $-6.803\text{eV}$ for $E_l$ in equation (III).

    $\begin{array}{c}\lambda =\frac{1239.8\text{eV}\cdot \text{nm}}{-1.701\text{eV}-\left(-6.803\text{eV}\right)}\\ =243.0\text{nm}\end{array}$

For $\lambda \left(K_{\beta }\right)$ : $n_l=1\text{and}{n}_u=3$.

Substitute $1239.8\text{eV}\cdot \text{nm}$ for $hc$, $-0.756\text{eV}$ for $E_u$ and $-6.803\text{eV}$ for $E_l$ in equation (III).

    $\begin{array}{c}\lambda =\frac{1239.8\text{eV}\cdot \text{nm}}{-0.756\text{eV}-\left(-6.803\text{eV}\right)}\\ =205.0\text{nm}\end{array}$

For $\lambda \left(L_{\alpha }\right)$ : $n_l=2\text{and}{n}_u=3$.

Substitute $1239.8\text{eV}\cdot \text{nm}$ for $hc$, $-0.756\text{eV}$ for $E_u$ and $-1.701\text{eV}$ for $E_l$ in equation (III).

    $\begin{array}{c}\lambda =\frac{1239.8\text{eV}\cdot \text{nm}}{-0.756\text{eV}-\left(-1.701\text{eV}\right)}\\ =1312\text{nm}\end{array}$

For $\lambda \left(L_{\beta }\right)$ : $n_l=2\text{and}{n}_u=4$.

Substitute $1239.8\text{eV}\cdot \text{nm}$ for $hc$, $-0.425\text{eV}$ for $E_u$ and $-1.701\text{eV}$ for $E_l$ in equation (III).

    $\begin{array}{c}\lambda =\frac{1239.8\text{eV}\cdot \text{nm}}{-0.425\text{eV}-\left(-1.701\text{eV}\right)}\\ =971.6\text{nm}\end{array}$

Thus, the wavelengths of the $K_{\alpha }$ transition is $243.0\text{nm}$ $,K_{\beta }$ transitions is $205.0\text{nm}$ $,L_{\alpha }$ transition is $1312\text{nm}$ and ${\text{L}}_{\beta }$ transition is $971.6\text{nm}$.