EBK LOOSE-LEAF VERSION OF UNIVERSE
EBK LOOSE-LEAF VERSION OF UNIVERSE
11th Edition
ISBN: 9781319227975
Author: KAUFMANN
Publisher: VST
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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.

(a)

Expert Solution
Check Mark

Answer to Problem 13Q

Solution:

1.6×1013 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=mc2

Here, E is the energy released, m is the total mass of the positron and electron, and c is the speed of light, 3×108 m/s.

Explanation:

The mass of the electron is 9.1×1031 kg and the mass of the positron is also 9.1×1031 kg.

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

m=(9.1×1031 kg)+(9.1×1031 kg)=1.82×1030 kg

Recall the expression for the mass-energy equation.

E=mc2

Substitute 1.82×1030 kg for m and 3×108 m/s for c.

E=(1.82×1030 kg)(3×108 m/s)2=(1.82×1030 kg)(9×1016 kg)=1.6×1013 J

Conclusion:

Therefore, the amount of energy released by the annihilation of electron and positron is 1.6×1013 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.

(b)

Expert Solution
Check Mark

Answer to Problem 13Q

Solution:

0.013 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=hcλ

Here, E is the energy received by each photon, λ is the wavelength of each photon, and h is Planck’s constant, 6.6×1034 J.s.

Explanation:

Each photon receives half of the total energy released in annihilation. So, the energy received by each photon is 8.1×1014 J.

Recall the expression for the relationship between energy and wavelength.

E=hcλ

Rearrange the above expression for λ.

λ=hcE

Substitute 8.1×1014 J for E, 6.6×1034 Js for h and 3×108 m/s for c.

λ=(6.6×1034 Js)(3×108 m/s)8.1×1013 J=2.4×1012 m(109 nm1 m)=2.4×103 nm

Refer to figure 5-7 from the textbook and determine whether the calculated wavelength 2.4×103 nm lies in the range of gamma-rays. The range of gamma rays is 0.01 nm to 10.0 nm.

Conclusion:

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

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