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Chapter 36: Problem 33

X-rays of wavelength \(\lambda=0.120 \mathrm{nm}\) are scattered from carbon. What is the Compton wavelength shift for photons detected at \(90.0^{\circ}\) angle relative to the incident beam?

Short Answer

Expert verified
Answer: The Compton wavelength shift for X-rays scattered at a 90-degree angle is 2.43 nm.

Step by step solution

01

Given information and Compton scattering formula

We are given the following information: - Initial wavelength of the X-rays, \(\lambda = 0.120 \, \mathrm{nm}\) - Scattering angle, \(\theta = 90.0^{\circ}\) The Compton scattering formula is given by: $$ \Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos{\theta}) $$ Here, \(h\) is the Planck's constant \((6.626 \times 10^{-34} \, \mathrm{J \cdot s})\), \(m_e\) is the electron's mass \((9.109 \times 10^{-31} \, \mathrm{kg})\), \(c\) is the speed of light \((2.998 \times 10^8 \, \mathrm{m/s})\), and \(\Delta \lambda\) is the Compton wavelength shift. We need to find the value of \(\Delta \lambda\).
02

Convert given wavelength to meters

Before we proceed with the calculation, let's convert the given wavelength to meters: $$ \lambda = 0.120 \, \mathrm{nm} \times \frac{1 \, \mathrm{m}}{10^9 \, \mathrm{nm}} = 1.20 \times 10^{-10} \, \mathrm{m} $$
03

Calculate the Compton wavelength using the formula

Now, let's substitute the given values into the Compton scattering formula and compute \(\Delta \lambda\): $$ \Delta \lambda = \frac{h}{m_e c} (1 - \cos{90.0^{\circ}}) $$ $$ \Delta \lambda = \frac{(6.626 \times 10^{-34}\, \mathrm{J\cdot s})}{(9.109 \times 10^{-31} \, \mathrm{kg})(2.998 \times 10^8 \, \mathrm{m/s})}(1 - \cos{90.0^{\circ}}) $$ Since \(\cos{90.0^{\circ}} = 0\), the formula simplifies to: $$ \Delta \lambda = \frac{h}{m_e c} = \frac{(6.626 \times 10^{-34}\, \mathrm{J\cdot s})}{(9.109 \times 10^{-31}\, \mathrm{kg})(2.998 \times 10^8\, \mathrm{m/s})} $$ Calculating the value, we get: $$ \Delta \lambda = 2.43 \times 10^{-12} \, \mathrm{m} $$
04

Convert the Compton wavelength shift to nanometers

Finally, let's convert the Compton wavelength shift to nanometers: $$ \Delta \lambda = 2.43 \times 10^{-12} \, \mathrm{m} \times \frac{10^9 \, \mathrm{nm}}{1 \, \mathrm{m}} = 2.43 \, \mathrm{nm} $$ The Compton wavelength shift for photons detected at a \(90.0^{\circ}\) angle relative to the incident beam is \(\boldsymbol{2.43\, \mathrm{nm}}\).

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Most popular questions from this chapter

Consider de Broglie waves for a Newtonian particle of mass \(m,\) momentum \(p=m v,\) and energy \(E=p^{2} /(2 m),\) that is, waves with wavelength \(\lambda=h / p\) and frequency \(f=E / h\). a) Calculate the dispersion relation \(\omega=\omega(k)\) for these waves. b) Calculate the phase and group velocities of these waves. Which of these corresponds to the classical velocity of the particle?

A nocturnal bird's eye can detect monochromatic light of frequency \(5.8 \cdot 10^{14} \mathrm{~Hz}\) with a power as small as \(2.333 \cdot 10^{-17} \mathrm{~W}\). What is the corresponding number of photons per second a nocturnal bird's eye can detect?

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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?

What would a classical physicist expect would be the result of shining a brighter UV lamp on a metal surface, in terms of the energy of emitted electrons? How does this differ from what the theory of the photoelectric effect predicts?
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