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.

Question

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Answer 1

Question

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

Answer to Problem 13Q

Solution:

1.6×10−13 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 9.1×10−31 kg and the mass of the positron is also 9.1×10−31 kg.

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

m=(9.1×10−31 kg)+(9.1×10−31 kg)=1.82×10−30 kg

Recall the expression for the mass-energy equation.

E=mc2

Substitute 1.82×10−30 kg for m and 3×108 m/s for c.

E=(1.82×10−30 kg)(3×108 m/s)2=(1.82×10−30 kg)(9×1016 kg)=1.6×10−13 J

Conclusion:

Therefore, the amount of energy released by the annihilation of electron and positron is 1.6×10−13 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

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, 6.6×10−34 J.s.

Explanation:

Each photon receives half of the total energy released in annihilation. So, the energy received by each photon is 8.1×10−14 J.

Recall the expression for the relationship between energy and wavelength.

E′=hcλ

Rearrange the above expression for λ.

λ=hcE′

Substitute 8.1×10−14 J for E′, 6.6×10−34 J⋅s for h and 3×108 m/s for c.

λ=(6.6×10−34 J⋅s)(3×108 m/s)8.1×10−13 J=2.4×10−12 m(109 nm1 m)=2.4×10−3 nm

Refer to figure 5-7 from the textbook and determine whether the calculated wavelength 2.4×10−3 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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Answer 2

Question

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

Answer to Problem 13Q

Solution:

1.6×10−13 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 9.1×10−31 kg and the mass of the positron is also 9.1×10−31 kg.

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

m=(9.1×10−31 kg)+(9.1×10−31 kg)=1.82×10−30 kg

Recall the expression for the mass-energy equation.

E=mc2

Substitute 1.82×10−30 kg for m and 3×108 m/s for c.

E=(1.82×10−30 kg)(3×108 m/s)2=(1.82×10−30 kg)(9×1016 kg)=1.6×10−13 J

Conclusion:

Therefore, the amount of energy released by the annihilation of electron and positron is 1.6×10−13 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

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, 6.6×10−34 J.s.

Explanation:

Each photon receives half of the total energy released in annihilation. So, the energy received by each photon is 8.1×10−14 J.

Recall the expression for the relationship between energy and wavelength.

E′=hcλ

Rearrange the above expression for λ.

λ=hcE′

Substitute 8.1×10−14 J for E′, 6.6×10−34 J⋅s for h and 3×108 m/s for c.

λ=(6.6×10−34 J⋅s)(3×108 m/s)8.1×10−13 J=2.4×10−12 m(109 nm1 m)=2.4×10−3 nm

Refer to figure 5-7 from the textbook and determine whether the calculated wavelength 2.4×10−3 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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Answer 3

Question

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

Answer to Problem 13Q

Solution:

1.6×10−13 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 9.1×10−31 kg and the mass of the positron is also 9.1×10−31 kg.

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

m=(9.1×10−31 kg)+(9.1×10−31 kg)=1.82×10−30 kg

Recall the expression for the mass-energy equation.

E=mc2

Substitute 1.82×10−30 kg for m and 3×108 m/s for c.

E=(1.82×10−30 kg)(3×108 m/s)2=(1.82×10−30 kg)(9×1016 kg)=1.6×10−13 J

Conclusion:

Therefore, the amount of energy released by the annihilation of electron and positron is 1.6×10−13 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

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, 6.6×10−34 J.s.

Explanation:

Each photon receives half of the total energy released in annihilation. So, the energy received by each photon is 8.1×10−14 J.

Recall the expression for the relationship between energy and wavelength.

E′=hcλ

Rearrange the above expression for λ.

λ=hcE′

Substitute 8.1×10−14 J for E′, 6.6×10−34 J⋅s for h and 3×108 m/s for c.

λ=(6.6×10−34 J⋅s)(3×108 m/s)8.1×10−13 J=2.4×10−12 m(109 nm1 m)=2.4×10−3 nm

Refer to figure 5-7 from the textbook and determine whether the calculated wavelength 2.4×10−3 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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Answer 4

Question

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

Answer to Problem 13Q

Solution:

1.6×10−13 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 9.1×10−31 kg and the mass of the positron is also 9.1×10−31 kg.

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

m=(9.1×10−31 kg)+(9.1×10−31 kg)=1.82×10−30 kg

Recall the expression for the mass-energy equation.

E=mc2

Substitute 1.82×10−30 kg for m and 3×108 m/s for c.

E=(1.82×10−30 kg)(3×108 m/s)2=(1.82×10−30 kg)(9×1016 kg)=1.6×10−13 J

Conclusion:

Therefore, the amount of energy released by the annihilation of electron and positron is 1.6×10−13 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

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, 6.6×10−34 J.s.

Explanation:

Each photon receives half of the total energy released in annihilation. So, the energy received by each photon is 8.1×10−14 J.

Recall the expression for the relationship between energy and wavelength.

E′=hcλ

Rearrange the above expression for λ.

λ=hcE′

Substitute 8.1×10−14 J for E′, 6.6×10−34 J⋅s for h and 3×108 m/s for c.

λ=(6.6×10−34 J⋅s)(3×108 m/s)8.1×10−13 J=2.4×10−12 m(109 nm1 m)=2.4×10−3 nm

Refer to figure 5-7 from the textbook and determine whether the calculated wavelength 2.4×10−3 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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Answer 5

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

Answer to Problem 13Q

Solution:

1.6×10−13 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 9.1×10−31 kg and the mass of the positron is also 9.1×10−31 kg.

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

m=(9.1×10−31 kg)+(9.1×10−31 kg)=1.82×10−30 kg

Recall the expression for the mass-energy equation.

E=mc2

Substitute 1.82×10−30 kg for m and 3×108 m/s for c.

E=(1.82×10−30 kg)(3×108 m/s)2=(1.82×10−30 kg)(9×1016 kg)=1.6×10−13 J

Conclusion:

Therefore, the amount of energy released by the annihilation of electron and positron is 1.6×10−13 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

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, 6.6×10−34 J.s.

Explanation:

Each photon receives half of the total energy released in annihilation. So, the energy received by each photon is 8.1×10−14 J.

Recall the expression for the relationship between energy and wavelength.

E′=hcλ

Rearrange the above expression for λ.

λ=hcE′

Substitute 8.1×10−14 J for E′, 6.6×10−34 J⋅s for h and 3×108 m/s for c.

λ=(6.6×10−34 J⋅s)(3×108 m/s)8.1×10−13 J=2.4×10−12 m(109 nm1 m)=2.4×10−3 nm

Refer to figure 5-7 from the textbook and determine whether the calculated wavelength 2.4×10−3 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.

Answer 6

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

Answer to Problem 13Q

Solution:

1.6×10−13 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 9.1×10−31 kg and the mass of the positron is also 9.1×10−31 kg.

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

m=(9.1×10−31 kg)+(9.1×10−31 kg)=1.82×10−30 kg

Recall the expression for the mass-energy equation.

E=mc2

Substitute 1.82×10−30 kg for m and 3×108 m/s for c.

E=(1.82×10−30 kg)(3×108 m/s)2=(1.82×10−30 kg)(9×1016 kg)=1.6×10−13 J

Conclusion:

Therefore, the amount of energy released by the annihilation of electron and positron is 1.6×10−13 J.

Answer 7

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 8

(a)

Expert Solution

Answer to Problem 13Q

Solution:

1.6×10−13 J

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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 9.1×10−31 kg and the mass of the positron is also 9.1×10−31 kg.

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

m=(9.1×10−31 kg)+(9.1×10−31 kg)=1.82×10−30 kg

Recall the expression for the mass-energy equation.

E=mc2

Substitute 1.82×10−30 kg for m and 3×108 m/s for c.

E=(1.82×10−30 kg)(3×108 m/s)2=(1.82×10−30 kg)(9×1016 kg)=1.6×10−13 J

Conclusion:

Therefore, the amount of energy released by the annihilation of electron and positron is 1.6×10−13 J.

Answer 13

(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

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, 6.6×10−34 J.s.

Explanation:

Each photon receives half of the total energy released in annihilation. So, the energy received by each photon is 8.1×10−14 J.

Recall the expression for the relationship between energy and wavelength.

E′=hcλ

Rearrange the above expression for λ.

λ=hcE′

Substitute 8.1×10−14 J for E′, 6.6×10−34 J⋅s for h and 3×108 m/s for c.

λ=(6.6×10−34 J⋅s)(3×108 m/s)8.1×10−13 J=2.4×10−12 m(109 nm1 m)=2.4×10−3 nm

Refer to figure 5-7 from the textbook and determine whether the calculated wavelength 2.4×10−3 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.

Answer 14

(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 15

(b)

Expert Solution

Answer to Problem 13Q

Solution:

0.013 arcsec

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

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, 6.6×10−34 J.s.

Explanation:

Each photon receives half of the total energy released in annihilation. So, the energy received by each photon is 8.1×10−14 J.

Recall the expression for the relationship between energy and wavelength.

E′=hcλ

Rearrange the above expression for λ.

λ=hcE′

Substitute 8.1×10−14 J for E′, 6.6×10−34 J⋅s for h and 3×108 m/s for c.

λ=(6.6×10−34 J⋅s)(3×108 m/s)8.1×10−13 J=2.4×10−12 m(109 nm1 m)=2.4×10−3 nm

Refer to figure 5-7 from the textbook and determine whether the calculated wavelength 2.4×10−3 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.