A photon with a momentum of 1.02 MeV/s, where c is the speed of light in

A photon with a momentum of 1.02 MeV/s, where c is the speed of light in vacuum, was scattered by a free electron at rest, as a result of which the photon momentum decreased to 0.255 MeV/s. It is necessary to determine the photon scattering angle.

To solve this problem it is necessary to use the laws of conservation of energy and momentum. Let the photon scattering angle relative to the original direction be equal to θ. Then the photon energy before scattering is equal to its energy after scattering:

Ephoton before = Ephoton after

hc/λ before = hc/λ after

where h is Planck's constant, λ is the wavelength of the photon before and after scattering.

To find the scattering angle θ, you can use the law of conservation of momentum:

pphoton before = pphoton after

where p is the photon momentum before and after scattering.

The photon momentum before scattering is equal to:

pphoton to = Ephoton to/c

The photon momentum after scattering is equal to:

pphoton after = Ephoton after/c

Thus, the photon scattering angle can be determined by the formula:

cos(θ) = 1 - (λ after/λ before) = 1 - (pphoton after/pphoton before)

Answer: The photon scattering angle is equal to [insert answer].

Photon with momentum 1.02 MeV/s

The product is available as a digital product

This product is a fascinating physics problem. In it, it is necessary to calculate the scattering angle of a photon with a momentum of 1.02 MeV/s on a free electron at rest. To solve the problem it is necessary to use the laws of conservation of energy and momentum.

• Format: digital product
• Russian language
• Difficulty: medium
• Required knowledge: physics

Cost: 50 rubles

This digital product is an interesting problem from the field of physics, which will allow you to test your knowledge and apply it in practice. In the problem, it is necessary to calculate the scattering angle of a photon with a momentum of 1.02 MeV/c on a free electron at rest, using the laws of conservation of energy and momentum. The product is available in the format of a digital product in Russian, the level of complexity is medium. The cost of this product is 50 rubles.

This product is a digital problem from the field of physics. In the problem, it is necessary to calculate the scattering angle of a photon with a momentum of 1.02 MeV/c on a free electron at rest, using the laws of conservation of energy and momentum.

To solve the problem, you must use the following formulas and laws:

1. Law of conservation of energy: the energy of a photon before scattering is equal to its energy after scattering: Ephoton before = Ephoton after, where E is the photon energy, h is Planck’s constant, λ is the wavelength of the photon before and after scattering.

2. Law of conservation of momentum: the momentum of the photon before scattering is equal to the momentum of the photon after scattering: pphoton before = pphoton after, where p is the momentum of the photon before and after scattering.

3. The photon momentum before scattering is equal to: pphoton up = Ephoton up/c, where c is the speed of light in vacuum.

4. The photon momentum after scattering is equal to: pphoton after = Ephoton after/s.

5. The photon scattering angle can be determined by the formula: cos(θ) = 1 - (λ after/λ before) = 1 - (pphoton after/pphoton before), where θ is the photon scattering angle, λ is the photon wavelength before and after scattering , p is the photon momentum before and after scattering.

Answer to the problem: the photon scattering angle is approximately 60 degrees.

If you have questions about solving a problem, you can contact the seller of this digital product for help.

This product is a fascinating physics problem that allows you to calculate the scattering angle of a photon with a momentum of 1.02 MeV/s on a free electron at rest. To solve the problem it is necessary to use the laws of conservation of energy and momentum.

A photon with a momentum of 1.02 MeV/s is scattered by a free electron at rest, as a result of which the photon momentum decreases to 0.255 MeV/s. Using the laws of conservation of energy and momentum, the photon scattering angle can be calculated.

To do this, you need to use the following formulas:

• the energy of a photon before scattering is equal to its energy after scattering: Ephoton before = Ephoton after, where E is the photon energy, h is Planck’s constant, λ is the wavelength of the photon before and after scattering: hc/λ before = hc/λ after.
• The momentum of the photon before scattering is equal to the momentum of the photon after scattering: pphoton before = pphoton after, where p is the momentum of the photon before and after scattering: pphoton = Ephoton/c, where c is the speed of light in vacuum.

Using these formulas, we can obtain an equation for finding the photon scattering angle: cos(θ) = 1 - (λ after/λ before) = 1 - (pphoton after/pphoton before).

Substituting the known values, we get: cos(θ) = 1 - (0.255/1.02) = 0.75.

Solving this equation, we get: θ = arccos(0.75) ≈ 41.4 degrees.

Answer: The photon scattering angle is approximately 41.4 degrees.

A detailed solution to the problem with a brief description of the conditions, formulas and laws used in the solution, the derivation of the calculation formula and the answer can be found in product number 50107. If you have any questions about the solution, you can ask for help.

***

Product description:

A photon with a momentum of 1.02 MeV/c, where c is the speed of light in vacuum, is an elementary particle of the electromagnetic field that was scattered by a free electron at rest. As a result of scattering, the photon momentum decreased to 0.255 MeV/s.

To determine the photon scattering angle, you can use the law of conservation of energy and momentum during the scattering process. In this case, you can use Compton's formulas to calculate the change in wavelength and photon scattering angle.

By solving problem 50107, you can get a detailed solution with a brief record of the conditions, formulas and laws used in the solution, the derivation of the calculation formula and the answer. If you have any questions about the solution, you can ask for help.

***

1. The 1.02 MeV/s photon is an excellent digital product for scientific research.
2. It is convenient to use the digital product Photon with a pulse of 1.02 MeV/s in experiments that require high measurement accuracy.
3. Digital product Photon with a pulse of 1.02 MeV/s allows you to obtain accurate data in experiments using X-ray radiation.
4. Thanks to the Photon with a pulse of 1.02 MeV/s, it is possible to study the properties of materials at the microlevel.
5. This digital product will help reduce experiment time and increase efficiency.
6. The 1.02 MeV/s photon is a reliable and precise tool for research in physics and materials science.
7. Purchasing a digital product Photon with a pulse of 1.02 MeV/s is a profitable investment in scientific research.

Peculiarities:

A photon with a momentum of 1.02 MeV/s is an excellent digital product for scientific research.

Convenient and easy-to-use digital product Photon with a momentum of 1.02 MeV/s.

A photon with a momentum of 1.02 MeV/s is a reliable and accurate instrument for measuring photon energy.

The digital product Photon with an impulse of 1.02 MeV / s has an excellent price-quality ratio.

A photon with a momentum of 1.02 MeV/s makes it possible to obtain accurate and reliable results in experiments.

Incredibly useful and convenient digital product Photon with a momentum of 1.02 MeV / s for scientific research.

Great functionality and ease of use are the excellent qualities of the digital product Photon with an impulse of 1.02 MeV / s.

High accuracy and reliability of measurements together with ease of use - this is a Photon with a momentum of 1.02 MeV / s.

A photon with a momentum of 1.02 MeV/s is a necessary and indispensable tool in scientific laboratories.

Excellent quality and ease of use are the main advantages of the digital product Photon with a momentum of 1.02 MeV/s.