. At what scattering angle is the wavelength shift in the Compton effect equal to the Compton wavelength?

The Compton scattering formula describes the relationship between the initial and scattered wavelengths of a photon after an elastic collision with an electron. The formula is given by: \[\lambda' - \lambda = \frac{h}{m_e c}(1 - \cos{\theta})\] where: - \(\lambda'\) is the scattered wavelength, - \(\lambda\) is the initial wavelength, - \(h\) is the Planck's constant, - \(m_e\) is the electron mass, - \(c\) is the speed of light, - \(\theta\) is the scattering angle. The Compton wavelength, denoted as \(\lambda_C\), is given by: \[\lambda_C = \frac{h}{m_e c}\] Our task is to find the scattering angle \(\theta\) at which the wavelength shift (\(\lambda' - \lambda\)) equals the Compton wavelength (\(\lambda_C\)).

As per the problem, the wavelength shift is equal to the Compton wavelength. Therefore, we can write: \[\lambda_C = \lambda' - \lambda\] Substitute the Compton scattering formula and the Compton wavelength formula into the equation: \[\frac{h}{m_e c} = \frac{h}{m_e c}(1 - \cos{\theta})\]

To find the scattering angle \(\theta\), we need to solve the equation obtained in Step 2. Divide both sides of the equation by \(\frac{h}{m_e c}\): \[1 = (1 - \cos{\theta})\] Now, we need to isolate \(\theta\) in this equation: \[\cos{\theta} = 1 - 1\] \[\cos{\theta} = 0\] Using the inverse cosine function (also called arccos): \[\theta = \arccos(0)\] The arccos of 0 is equal to \(\frac{\pi}{2}\): \[\theta = \frac{\pi}{2}\]

In most cases, we express angles in degrees. To convert the result from radians to degrees, we use the formula: Degrees = Radians * \(\frac{180}{\pi}\) Substitute our result for \(\theta\) into this formula: Degrees = \(\frac{\pi}{2}\) * \(\frac{180}{\pi}\) Degrees = 90° The scattering angle at which the wavelength shift in the Compton effect is equal to the Compton wavelength is 90°.