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Chapter 27: Problem 71

Nuclei in a radium-226 radioactive source emit photons whose energy is $186 \mathrm{keV} .$ These photons are scattered by the electrons in a metal target; a detector measures the energy of the scattered photons as a function of the angle \(\theta\) through which they are scattered. Find the energy of the \(\gamma\) -rays scattered through \(\theta=90.0^{\circ}\) and \(180.0^{\circ} .\)

Short Answer

Expert verified
Answer: The scattered energies of the gamma-rays are 165.84 keV at \(90^{\circ}\) and 107.61 keV at \(180^{\circ}\).

Step by step solution

01

Recall the Compton Scattering formula

The Compton Scattering formula relates the initial energy of a photon \(E_{initial}\), the energy of a scattered photon \(E_{scattered}\), and the angle \(\theta\) it is scattered through as follows: $$E_{scattered} = \frac{E_{initial}}{1 + \frac{E_{initial}}{m_e c^2} (1 - \cos{\theta})}$$ Here, \(m_e\) is the mass of an electron, and \(c\) is the speed of light.
02

Convert the initial energy to Joules

The initial energy of the photon is given in keV. To use this energy in the Compton Scattering formula, we must convert it to Joules. 1 eV = 1.60218 × 10^-19 J So, 186 keV = 186 × 10^3 × 1.60218 × 10^-19 J = 2.97925 × 10^-14 J
03

Calculate the scattered energy for the first angle

We are given the first angle, \(\theta_1 = 90^{\circ}\). Plug in all the values and calculate the scattered energy for the first angle using the Compton Scattering formula: $$E_{scattered1} = \frac{2.97925 \times 10^{-14}}{1 + \frac{2.97925 \times 10^{-14}}{9.109 \times 10^{-31} \times 2.998 \times 10^8} (1 - \cos{90^{\circ}})}$$ Calculate the scattered energy: $$E_{scattered1} = 2.65715 \times 10^{-14} \mathrm{J}$$
04

Calculate the scattered energy for the second angle

We are given the second angle, \(\theta_2 = 180^{\circ}\). Plug in all the values and calculate the scattered energy for the second angle using the Compton Scattering formula: $$E_{scattered2} = \frac{2.97925 \times 10^{-14}}{1 + \frac{2.97925 \times 10^{-14}}{9.109 \times 10^{-31} \times 2.998 \times 10^8} (1 - \cos{180^{\circ}})}$$ Calculate the scattered energy: $$E_{scattered2} = 1.72415 \times 10^{-14} \mathrm{J}$$
05

Convert the scattered energies back to keV

To report our answer in keV, convert the scattered energies back to keV: Scattered energy at \(\theta = 90^{\circ}\) $$E_{scattered1} = \frac{2.65715 \times 10^{-14}}{1.60218 \times 10^{-19}} \mathrm{keV} = 165.84 \mathrm{keV}$$ Scattered energy at \(\theta = 180^{\circ}\) $$E_{scattered2} = \frac{1.72415 \times 10^{-14}}{1.60218 \times 10^{-19}} \mathrm{keV} = 107.61 \mathrm{keV}$$ The scattered energies for the given angles are: \(E_{scattered1} = 165.84 \mathrm{keV}\) at \(\theta=90^{\circ}\) and \(E_{scattered2} = 107.61 \mathrm{keV}\) at \(\theta=180^{\circ}\).

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Most popular questions from this chapter

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?
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