In 1923 Arthur Compton observed that \(x\)-rays scattered off free electrons were shifted in wavelength by an amount that could be explained by assuming that the \(\mathrm{x}\)-rays and electrons obeyed the relativistic relationships for energy and momentum. Compton used \(\mathrm{x}\)-rays \((\gamma)\) with a wavelength of approximately \(7.30 \times 10^{-11} \mathrm{~m}\). Assume that such a photon is incident on a stationary electron, as shown below. For purposes of illustration, the photon is reflected off the electron and its wavelength is observed to shift by a magnitude \(|\Delta \lambda|=\left|\lambda_o-\lambda_f\right|=h / m_e c\), where \(\lambda_o\) and \(\lambda_f\) are the initial and final wavelengths, respectively. a) What is the energy of the incoming photon in \(\mathrm{eV}\) ? in joules? b) Does the wavelength of the photon increase or decrease? Explain your reasoning. c) What is the momentum acquired by the electron?

Step by step solution

01

Calculate the photon's frequency (use energy-wavelength relationship and Planck's constant)

First, we'll use the energy-wavelength relationship and Planck's constant to calculate the frequency of the photon: \(\nu = \frac{c}{\lambda} = \frac{3.00\times 10^8 \mathrm{~m/s}}{7.30\times 10^{-11} \mathrm{~m}} = 4.11 \times 10^{18} \mathrm{~Hz}\)

02

Calculate the photon's energy in Joules (using energy-frequency relationship and Planck's constant)

Now, we'll use the energy-frequency relationship and Planck's constant to calculate the energy in Joules: \(E = h\nu = (6.63\times 10^{-34}\mathrm{~Js}) (4.11 \times 10^{18} \mathrm{~Hz}) = 2.72\times 10^{-15} \mathrm{~J}\)

03

Calculate the photon's energy in eV (divide energy in Joules by electric charge)

Finally, we'll convert the energy in Joules to electron volts (eV) by dividing by the electric charge: \(E_eV = \frac{2.72\times 10^{-15} \mathrm{~J}}{1.60\times 10^{-19} \mathrm{~C}} = 17,000 \mathrm{~eV}\) Thus, the energy of the incoming photon is \(17,000 \mathrm{~eV}\) or \(2.72\times 10^{-15} \mathrm{~J}\). #b) Discuss the change in wavelength#

04

Compare the initial and final wavelengths (using the given value for the change in wavelength)

We are given the magnitude of the shift in wavelength: \(|\Delta \lambda| = h / m_e c\). Since \(h / m_e c\) is positive, we can say that the final wavelength (\(\lambda_f\)) is greater than the initial wavelength (\(\lambda_o\)). As the wavelength of the photon increases, the energy of the photon decreases due to the inverse relationship between energy and wavelength. This decrease in energy is transferred to the electron, causing it to gain kinetic energy and momentum. #c) Calculate the momentum acquired by the electron#

05

Calculate the change in photon's momentum (use energy-momentum relationship)

The change in the photon's momentum (\(\Delta p_\gamma\)) can be calculated using the energy-momentum relationship: \(\Delta p_\gamma = \frac{\Delta E_\gamma}{c} = \frac{h (\frac{1}{\lambda_o} - \frac{1}{\lambda_f})}{c}\)

06

Calculate the momentum acquired by the electron (conservation of momentum)

Since momentum is conserved, the electron acquires the same amount of momentum as the amount lost by the x-ray photon: \(p_e = \Delta p_\gamma = \frac{h (\frac{1}{\lambda_o} - \frac{1}{\lambda_f})}{c} = \frac{h (\frac{1}{\lambda_o} - \frac{1}{\lambda_o + h/m_e c})}{c}\) Now, plugging in the values, we can compute the electron's momentum: \(p_e = \frac{(6.63 \times 10^{-34} \mathrm{~Js})(\frac{1}{7.30 \times 10^{-11} \mathrm{~m}} - \frac{1}{7.30 \times 10^{-11} \mathrm{~m} + \frac{6.63 \times 10^{-34} \mathrm{~Js}}{9.11 \times 10^{-31} \mathrm{~kg} \cdot 3.00 \times 10^8 \mathrm{~m/s}}})}{3.00 \times 10^8 \mathrm{~m/s}} = 2.44 \times 10^{-24} \mathrm{~kg \cdot m/s}\) Thus, the momentum acquired by the electron is \(2.44 \times 10^{-24} \mathrm{~kg \cdot m/s}\).

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