Calculate the following energies: the scattered gamma ray following a Compton interaction of a 140 keV gamma ray in which the scattering angle 0 is 45°.
Calculate the following energies: the scattered gamma ray following a Compton interaction of a 140 keV gamma ray in which the scattering angle 0 is 45°.
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Question
![**Problem Statement:**
Calculate the following energies:
Compute the energy of the scattered gamma ray following a Compton interaction of a 140 keV gamma ray when the scattering angle θ is 45°.
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**Explanation:**
This exercise involves the Compton effect, a phenomenon in which a gamma ray interacts with a target, resulting in some of its energy being transferred and the gamma ray being scattered. Given the initial energy of the gamma ray (140 keV) and the angle of scattering (45°), the task is to determine the energy of the scattered gamma ray.
To solve this, the Compton wavelength shift equation and relevant formulas are used:
\[ \frac{1}{E'} = \frac{1}{E} + \frac{1}{m_ec^2}(1 - \cos \theta) \]
Where:
- \( E' \) is the energy of the scattered photon.
- \( E \) is the initial energy of the photon (140 keV).
- \( m_ec^2 \) is the rest energy of the electron (511 keV).
- \( \theta \) is the scattering angle (45°).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F203b58df-0ca9-47a6-a660-754b46a1e0bb%2Febbb2b0d-a9f3-4628-92ae-ab788f6177da%2Fam0c4au_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Calculate the following energies:
Compute the energy of the scattered gamma ray following a Compton interaction of a 140 keV gamma ray when the scattering angle θ is 45°.
---
**Explanation:**
This exercise involves the Compton effect, a phenomenon in which a gamma ray interacts with a target, resulting in some of its energy being transferred and the gamma ray being scattered. Given the initial energy of the gamma ray (140 keV) and the angle of scattering (45°), the task is to determine the energy of the scattered gamma ray.
To solve this, the Compton wavelength shift equation and relevant formulas are used:
\[ \frac{1}{E'} = \frac{1}{E} + \frac{1}{m_ec^2}(1 - \cos \theta) \]
Where:
- \( E' \) is the energy of the scattered photon.
- \( E \) is the initial energy of the photon (140 keV).
- \( m_ec^2 \) is the rest energy of the electron (511 keV).
- \( \theta \) is the scattering angle (45°).
Expert Solution
Step 1
The incident energy is
The energy is
where h is the Planck's constant, c is the speed of light and is the incident wavelength
The incident wavelength becomes
The value of is
In Compton scattering process, the difference of final and incident wavelength is
.............................. (1)
where is the scattering angle.
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