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Chapter 36: Problem 70

Compton used photons of wavelength \(0.0711 \mathrm{nm} .\) a) What is the wavelength of the photons scattered at \(\theta=180 .\) ? b) What is energy of these photons? c) If the target were a proton and not an electron, how would your answer in (a) change?

Short Answer

Expert verified
Answer: The scattered photon's wavelength is approximately 0.0821 nm, its energy is approximately 2.41 x 10^(-16) J, and the scattered wavelength would be approximately 0.0711 nm if the target were a proton.

Step by step solution

01

Calculate the wavelength of the scattered photon

To find the wavelength of the scattered photon, we will use the Compton scattering formula: \(\lambda' = \lambda + \frac{h}{m_ec}(1-\cos\theta)\) Where \(\lambda' = \) scattered wavelength, \(\lambda = 0.0711\,\mathrm{nm}\) (initial wavelength) \(h = 6.626 \times 10^{-34}\,\mathrm{J}\cdot\mathrm{s}\) (Planck's constant) \(m_e = 9.109 \times 10^{-31}\,\mathrm{kg}\) (mass of the electron) \(c = 3.0 \times 10^8\,\mathrm{m/s}\) (speed of light) \(\theta = 180^\circ\) (scattering angle)
02

Convert the angle from degrees to radians

To find the cosine of the angle, we first need to convert the given angle from degrees to radians: \(\theta_\mathrm{rad} = \frac{180}{\pi}\) radians
03

Calculate the scattered wavelength

Now, we can calculate the scattered wavelength using the Compton scattering formula: \(\lambda' = 0.0711\,\mathrm{nm} + \frac{6.626 \times 10^{-34}\,\mathrm{J}\cdot\mathrm{s}}{9.109 \times 10^{-31}\,\mathrm{kg} \cdot 3.0 \times 10^8\,\mathrm{m/s}}(1-\cos(180/\pi))\) \(\lambda' \approx 0.0821\,\mathrm{nm}\)
04

Calculate the energy of the scattered photons

The energy of the scattered photon can be found using the Planck's equation: \(E = \frac{hc}{\lambda'}\) Where \(E\) is the energy of the scattered photon \(h\) is the Planck's constant \(c\) is the speed of light \(\lambda'\) is the scattered wavelength Let's use the values: \(E = \frac{6.626 \times 10^{-34}\,\mathrm{J}\cdot\mathrm{s}\cdot 3.0 \times 10^8\,\mathrm{m/s}}{0.0821 \times 10^{-9}\,\mathrm{m}} \approx 2.41 \times 10^{-16}\,\mathrm{J}\)
05

Calculate the scattered photon's wavelength if the target were a proton

For this part, we need to replace the mass of the electron with the mass of the proton in the Compton scattering formula. The mass of the proton is given as: \(m_p = 1.673 \times 10^{-27}\,\mathrm{kg}\) Now, let's repeat step 3: \(\lambda'_p = 0.0711\,\mathrm{nm} + \frac{6.626 \times 10^{-34}\,\mathrm{J}\cdot\mathrm{s}}{1.673 \times 10^{-27}\,\mathrm{kg} \cdot 3.0 \times 10^8\,\mathrm{m/s}}(1-\cos(180/\pi))\) \(\lambda'_p \approx 0.0711\,\mathrm{nm}\) Since \(\lambda'_p \approx \lambda\), the change in the scattered wavelength will be negligible if the target was a proton instead of an electron.

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Most popular questions from this chapter

Consider an electron whose de Broglie wavelength is equal to the wavelength of green light (about \(550 \mathrm{nm}\) ). a) Treating the electron nonrelativistically, what is its speed? b) Does your calculation confirm that a nonrelativistic treatment is sufficient? c) Calculate the kinetic energy of the electron in \(\mathrm{eV}\).

What is the maximum kinetic energy of the electrons ejected from a sodium surface by light of wavelength \(470 \mathrm{nm} ?\)

Consider the equivalent of Compton scattering, but the case in which a photon scatters off of a free proton. a) If \(140 .-\mathrm{keV} \mathrm{X}\) -rays bounce off of a proton at \(90.0^{\circ},\) what is their fractional change in energy \(\left(E_{0}-E\right) / E_{0} ?\) b) What energy of photon would be necessary to cause a \(1.00 \%\) change in energy at \(90.0^{\circ}\) scattering?

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