Question

A thin aluminum target is illuminated with photons of wavelength \(\lambda\). A detector is placed at \(90.0^{\circ}\) to the direction of the incident photons. The scattered photons detected are found to have half the energy of the incident photons. (a) Find \(\lambda\). (b) What is the wavelength of backscattered photons (detector at \(\left.180^{\circ}\right) ?\) (c) What (if anything) would change if a copper target were used instead of an aluminum one?

Short answer

The incident wavelength is approximately \(\lambda \approx 2.43 \times 10^{-12}\,\text{m}\). b) What is the wavelength of the backscattered photons? The wavelength of the backscattered photons is \(\lambda_{180}' = 7.29 \times 10^{-12}\,\text{m}\). c) How would the scattering change if a copper target were used instead of aluminum? There would be no change in the scattering pattern if a copper target were used instead of aluminum because the Compton scattering formula does not depend on material properties.

Step-by-step solution

a) Finding the incident wavelength \(\lambda\)

Using the Compton scattering formula, the change in the scattered photon wavelength is given by: \(\Delta \lambda = \lambda' - \lambda = \dfrac{h}{m_e c}(1 - \cos\theta)\) , where \(\lambda'\) is the scattered photon wavelength, \(\theta\) is the scattering angle, \(h\) is the Planck's constant, \(= 6.63 × 10^{-34}\,\text{Js}\), \(m_e\) is the electron mass, \(= 9.11 × 10^{-31}\,\text{kg}\) \(c\) is the speed of light, \(= 3 \times 10^8\,\text{m/s}\). We are given that the scattered photon energy is half the incident photon energy, so \(\dfrac{hc}{\lambda'} = \dfrac{1}{2}\dfrac{hc}{\lambda}\). Rearranging for the ratio of wavelengths, we have \(\dfrac{\lambda'}{\lambda} = 2\). At a scattering angle of \(90^\circ\), we plug in the values: \(\Delta \lambda = \dfrac{h}{m_e c}(1 - \cos{90^\circ}) = \dfrac{h}{m_e c}\). So, \(\lambda' = 2\lambda = \lambda + \dfrac{h}{m_e c}\). Now we can find the incident wavelength \(\lambda\): \(\lambda = \dfrac{h}{m_e c} = \dfrac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 3 \times 10^8} \,\text{m} \approx 2.43 \times 10^{-12}\,\text{m}\).

b) Finding the wavelength of the backscattered photons

We now need to find the wavelength for a scattering angle of \(180^\circ\): \(\Delta \lambda_{180} = \dfrac{h}{m_e c}(1 - \cos{180^\circ}) = 2\dfrac{h}{m_e c}\). Thus, the backscattered wavelength \(\lambda_{180}' = \lambda + 2\dfrac{h}{m_e c} = 2\lambda + \dfrac{h}{m_e c}\). Plugging in the values for \(\lambda\), we get \(\lambda_{180}' = 2(2.43 \times 10^{-12}) + 2.43 \times 10^{-12} \,\text{m} = 7.29 \times 10^{-12}\,\text{m}\).

c) Change in the scattering if a copper target were used instead of aluminum

In the Compton scattering formula, there are no material-dependent factors. The scattering depends on the incident photon wavelength, scattering angle, and electron properties. As a result, there would be no change in the scattering pattern if a copper target were used instead of aluminum.