X-rays with a wavelength of 120.0 pm undergo Compton scattering. Find the wavelengths of the photons scattered at angle of 30.0°?
X-rays with a wavelength of 120.0 pm undergo Compton scattering. Find the wavelengths of the photons scattered at angle of 30.0°?
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![### Compton Scattering of X-rays
**Problem Statement:**
X-rays with a wavelength of 120.0 pm undergo Compton scattering. Find the wavelengths of the photons scattered at an angle of 30.0°.
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**Explanation:**
Compton scattering is a quantum mechanical phenomenon where X-ray or gamma-ray photons are scattered by electrons, resulting in an increase in the wavelength of the scattered photons. The change in the photon's wavelength (Δλ) depends on the scattering angle (θ) and can be calculated using the Compton wavelength shift formula:
\[ \Delta \lambda = \lambda_{f} - \lambda_{i} = \frac{h}{m_e c} (1 - \cos \theta) \]
Where:
- \( \lambda_{i} \) = initial wavelength of the photon
- \( \lambda_{f} \) = final wavelength of the photon
- \( h \) = Planck's constant (6.626 x 10^-34 Js)
- \( m_e \) = mass of the electron (9.11 x 10^-31 kg)
- \( c \) = speed of light (3 x 10^8 m/s)
- \( \theta \) = scattering angle
**Given Data:**
- Initial wavelength (\( \lambda_{i} \)) = 120.0 pm
- Scattering angle (\( \theta \)) = 30.0°
**Calculations:**
First, compute the Compton wavelength shift:
\[ \Delta \lambda = \frac{h}{m_e c} (1 - \cos 30.0^\circ) \]
\[ \Delta \lambda = \frac{6.626 \times 10^{-34} \text{ Js}}{9.11 \times 10^{-31} \text{ kg} \times 3 \times 10^8 \text{ m/s}} (1 - \cos 30.0^\circ) \]
\[ \Delta \lambda = 2.43 \times 10^{-12} \text{ m} \times (1 - \cos 30.0^\circ) \]
Using the cosine of 30.0°:
\[ \cos 30.0^\circ = \frac{\sqrt{3}}{2} \]
\[ \Delta \lambda = 2.43 \times 10^{-12} \text](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3fe8677b-d2b4-4cf1-b1ed-08820154fcb5%2Fdf33d95b-29d8-4100-9e54-7db99aaf1455%2Flyt6xe_processed.png&w=3840&q=75)
Transcribed Image Text:### Compton Scattering of X-rays
**Problem Statement:**
X-rays with a wavelength of 120.0 pm undergo Compton scattering. Find the wavelengths of the photons scattered at an angle of 30.0°.
---
**Explanation:**
Compton scattering is a quantum mechanical phenomenon where X-ray or gamma-ray photons are scattered by electrons, resulting in an increase in the wavelength of the scattered photons. The change in the photon's wavelength (Δλ) depends on the scattering angle (θ) and can be calculated using the Compton wavelength shift formula:
\[ \Delta \lambda = \lambda_{f} - \lambda_{i} = \frac{h}{m_e c} (1 - \cos \theta) \]
Where:
- \( \lambda_{i} \) = initial wavelength of the photon
- \( \lambda_{f} \) = final wavelength of the photon
- \( h \) = Planck's constant (6.626 x 10^-34 Js)
- \( m_e \) = mass of the electron (9.11 x 10^-31 kg)
- \( c \) = speed of light (3 x 10^8 m/s)
- \( \theta \) = scattering angle
**Given Data:**
- Initial wavelength (\( \lambda_{i} \)) = 120.0 pm
- Scattering angle (\( \theta \)) = 30.0°
**Calculations:**
First, compute the Compton wavelength shift:
\[ \Delta \lambda = \frac{h}{m_e c} (1 - \cos 30.0^\circ) \]
\[ \Delta \lambda = \frac{6.626 \times 10^{-34} \text{ Js}}{9.11 \times 10^{-31} \text{ kg} \times 3 \times 10^8 \text{ m/s}} (1 - \cos 30.0^\circ) \]
\[ \Delta \lambda = 2.43 \times 10^{-12} \text{ m} \times (1 - \cos 30.0^\circ) \]
Using the cosine of 30.0°:
\[ \cos 30.0^\circ = \frac{\sqrt{3}}{2} \]
\[ \Delta \lambda = 2.43 \times 10^{-12} \text
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