Chapter 16, Problem 13Q

(a)

To determine

The amount of the energy released by the annihilation of an electron and a positron, if a positron has the same mass as an electron.

Answer to Problem 13Q

Solution:

$1.6\times {10}^{-13}\text{ J}$

Explanation of Solution

Given data:

The mass of the electron and the positron is the same.

Formula used:

According to the mass energy equation, the relation between energy and mass is:

$E=m{c}^2$

Here, E is the energy released, m is the total mass of the positron and electron, and c is the speed of light, $3\times {10}^8\text{ m/s}$.

Explanation:

The mass of the electron is $9.1\times {10}^{-31}\text{ kg}$ and the mass of the positron is also $9.1\times {10}^{-31}\text{ kg}$.

Then, the total mass of the electron and positron is,

$\begin{array}{c}m=\left(9.1\times {10}^{-31}\text{ kg}\right)+\left(9.1\times {10}^{-31}\text{ kg}\right)\\ \text{=1}\text{.82}\times {\text{10}}^{-30}\text{ kg}\end{array}$

Recall the expression for the mass-energy equation.

$E=m{c}^2$

Substitute $1.82\times {10}^{-30}\text{ kg}$ for m and $3\times {10}^8\text{ m/s}$ for c.

$\begin{array}{c}E=\left(1.82\times {10}^{-30}\text{ kg}\right){\left(3\times {10}^8\text{ m/s}\right)}^2\\ =\left(1.82\times {10}^{-30}\text{ kg}\right)\left(9\times {10}^{16}\text{ kg}\right)\\ =1.6\times {10}^{-13}\text{ J}\end{array}$

Conclusion:

Therefore, the amount of energy released by the annihilation of electron and positron is $1.6\times {10}^{-13}\text{ J}$.

(b)

To determine

The wavelength of each photon. Also, confirm from figure 5-7 (from textbook) that this wavelength falls in the range of gamma-rays, if the products of the annihilation are two photons each of equal energy.

Answer to Problem 13Q

Solution:

$0.013\text{ arcsec}$

Explanation of Solution

Given data:

The products of the annihilation are two photons, each of equal energy.

Formula used:

The expression for the relationship between energy and wavelength is:

$E^{\prime }=\frac{hc}{\lambda }$

Here, $E^{\prime }$ is the energy received by each photon, $\lambda$ is the wavelength of each photon, and h is Planck’s constant, $6.6\times {10}^{-34}\text{ J}\text{.s}$.

Explanation:

Each photon receives half of the total energy released in annihilation. So, the energy received by each photon is $8.1\times {10}^{-14}\text{ J}$.

Recall the expression for the relationship between energy and wavelength.

$E^{\prime }=\frac{hc}{\lambda }$

Rearrange the above expression for $\lambda$.

$\lambda =\frac{hc}{E^{\prime }}$

Substitute $8.1\times {10}^{-14}\text{ J}$ for $E^{\prime }$, $6.6\times {10}^{-34}\text{ J}\cdot \text{s}$ for h and $3\times {10}^8\text{ m/s}$ for c.

$\begin{array}{c}\lambda =\frac{\left(6.6\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3\times {10}^8\text{ m/s}\right)}{8.1\times {10}^{-13}\text{ J}}\\ =2.4\times {10}^{-12}\text{ m}\left(\frac{{10}^9\text{ nm}}{1\text{ m}}\right)\\ =2.4\times {10}^{-3}\text{ nm}\end{array}$

Refer to figure 5-7 from the textbook and determine whether the calculated wavelength $2.4\times {10}^{-3}\text{ nm}$ lies in the range of gamma-rays. The range of gamma rays is $0.01\text{ nm}$ to $\text{10}\text{.0 nm}$.

Conclusion:

Therefore, this wavelength lies in the range of gamma-rays.