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

Want to see more full solutions like this?

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Suppose a star 1000 times brighter than our Sun (that is, emitting 1000 times the power) suddenly goes supernova. Using data from Table: (a) By what factor does its power output increase? (b) How many times brighter than our entire Milky Way galaxy is the supernova? (c) Based on your answers, discuss whether it should be possible to observe supernovas in distant galaxies. Note that there are on the order of 1011 observable galaxies, the average brightness of which is somewhat less than our own galaxy.

suppose you collected a data set in which you measured fall-times for different fall-heights, you plotted the data and fit the mathemathical model, y=Ax^2, to match the physical hypothesis, y=0.5*g*t^2. From the best fit curve you are told that the value of your fit-parameter, A, is for 4.6 m/s^2 +/- 0.4 m/s^2. Determine the value of g +/- delta g for this fit parameter value.

Problem 2: Black hole – the ultimate blackbody A black hole emits blackbody radiation called Hawking radiation. A black hole with mass M has a total energy of Mc², a surface area of 167G²M² /c*, and a temperature of hc³/167²KGM. a) Estimate the typical wavelength of the Hawking radiation emitted by a 1 solar mass black hole (2 × 103ºkg). Compare your answer to the size of the black hole. b) Calculate the total power radiated by a one-solar mass black hole. c) Imagine a black hole in empty space, where it emits radiation but absorbs nothing. As it loses energy, its mass must decrease; one could say "evaporates". Derive a differential equation for the mass as a function of time, and solve to obtain an expression for the lifetime of a black hole in terms of its mass.

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.