Answer: [tex]90\°[/tex]
Explanation:
The Compton Shift [tex]\Delta \lambda[/tex] in wavelength when the photons are scattered is given by the following equation:
[tex]\Delta \lambda=\lambda_{c}(1-cos\theta)[/tex] (1)
Where:
[tex]\lambda_{c}=2.43(10)^{-12} m[/tex] is a constant whose value is given by [tex]\frac{h}{m_{e}c}[/tex], being [tex]h[/tex] the Planck constant, [tex]m_{e}[/tex] the mass of the electron and [tex]c[/tex] the speed of light in vacuum.
[tex]\theta)[/tex] the angle between incident phhoton and the scatered photon.
We are told the maximum Compton shift in wavelength occurs when a photon isscattered through [tex]180\°[/tex]:
[tex]\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))[/tex] (2)
[tex]\Delta \lambda_{max}=\lambda_{c}(1-(-1))[/tex]
[tex]\Delta \lambda_{max}=2\lambda_{c}[/tex] (3)
Now, let's find the angle that will produce a fourth of this maximum value found in (3):
[tex]\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)[/tex] (4)
[tex]\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)[/tex] (5)
If we want [tex]\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}[/tex], [tex]1-cos\theta[/tex] must be equal to 1:
[tex]1-cos\theta=1[/tex] (6)
Finding [tex]\theta[/tex]:
[tex]1-1=cos\theta[/tex]
[tex]0=cos\theta[/tex]
[tex]\theta=cos^{-1} (0)[/tex]
Finally:
[tex]\theta=90\°[/tex] This is the scattering angle that will produce [tex]\frac{1}{4}\Delta \lambda_{max}[/tex]
