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Chapter 6: Problem 159

Compare the wavelength shift of a photon scattered by a free electron to that of a photon scattered at the same angle by a free proton.

Short Answer

Expert verified
The wavelength shift of a photon scattered by a free electron is approximately 1840 times larger than that of a photon scattered by a free proton at the same angle. This difference is due to the mass of the scattering particles, with the electron being significantly lighter than the proton.

Step by step solution

01

Recall the Compton Scattering Formula

The Compton scattering formula can be expressed as: \(\Delta \lambda = \dfrac{h}{m_ec}(1-\cos{\theta})\), where \(\Delta \lambda\) is the wavelength shift, \(h\) is the Planck's constant, \(m_e\) is the mass of the scattering particle (electron or proton in this case), \(c\) is the speed of light, and \(\theta\) is the angle between the scattered photon and the incident photon.
02

Calculate the wavelength shift for the electron

Let's first calculate the wavelength shift for the electron. We need to plug in the known constant values and the mass of an electron into the Compton scattering formula: \(m_e = 9.11 \times 10^{-31} \, kg\) \(\Delta \lambda_e = \dfrac{6.63 \times 10^{-34} Js}{9.11 \times 10^{-31} kg \times 3 \times 10^8 ms^{-1}} (1-\cos{\theta})\) \(\Delta \lambda_e = \dfrac{2.43 \times 10^{-12}}{1 - \cos{\theta}}\, m\)
03

Calculate the wavelength shift for the proton

Now let's calculate the wavelength shift for the proton. We need to plug in the known constant values and the mass of a proton into the Compton scattering formula: \(m_p = 1.67 \times 10^{-27} \, kg\) \(\Delta \lambda_p = \dfrac{6.63 \times 10^{-34} Js}{1.67 \times 10^{-27} kg \times 3 \times 10^8 ms^{-1}} (1-\cos{\theta})\) \(\Delta \lambda_p = \dfrac{1.32 \times 10^{-15}}{1 - \cos{\theta}}\, m\)
04

Compare the wavelength shifts

To compare the wavelength shifts, we can divide the shift for the electron by the shift for the proton: \(\dfrac{\Delta \lambda_e}{\Delta \lambda_p} = \dfrac{\dfrac{2.43 \times 10^{-12}}{1 - \cos{\theta}}}{\dfrac{1.32 \times 10^{-15}}{1 - \cos{\theta}}}\) After canceling the denominator, we are left with: \(\dfrac{\Delta \lambda_e}{\Delta \lambda_p} = \dfrac{2.43 \times 10^{-12}}{1.32 \times 10^{-15}}\) \(\dfrac{\Delta \lambda_e}{\Delta \lambda_p} \approx 1840\) So, the wavelength shift of a photon scattered by a free electron is approximately 1840 times larger than that of a photon scattered by a free proton at the same angle.

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