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Chapter 28: Problem 7

An x-ray diffraction experiment using 16 -keV x-rays is repeated using electrons instead of \(x\) -rays. What should the kinetic energy of the electrons be in order to produce the same diffraction pattern as the x-rays (using the same crystal)?

Short Answer

Expert verified
Answer: To find the kinetic energy of electrons required to produce the same diffraction pattern as the x-rays, follow these steps: 1. Calculate the x-ray wavelength using the energy-wavelength relationship: \(\lambda_x = \frac{hc}{E_x}\) 2. Find the de Broglie wavelength of electrons with the same wavelength as the x-rays (\(\lambda_e = \lambda_x\)). 3. Determine the electron momentum using the de Broglie wavelength formula: \(p_e = \frac{h}{\lambda_e}\) 4. Relate electron momentum to its kinetic energy: \(K_e = \frac{p_e^2}{2m_e}\) 5. Calculate the kinetic energy of electrons in Joules and convert it to electron volts (eV): \(K_e (eV) = \frac{K_e (J)}{1.6 \times 10^{-19}\text{J/eV}}\) By following these steps and plugging in the given values and known constants, you can find the kinetic energy of electrons required to produce the same diffraction pattern as 16-keV x-rays using the same crystal.

Step by step solution

01

Calculate the x-ray wavelength

First, convert the x-ray energy to the wavelength. We are given the x-ray energy (\(E_x\)) as 16 keV, which needs to be converted to Joules (1 eV = 1.6 x 10^{-19} J). The relationship between an x-ray's energy and its wavelength (\(\lambda_x\)) is: \(\lambda_x = \frac{hc}{E_x}\) where \(h\) is the Planck's constant (\(6.626 \times 10^{-34}\) Js), and \(c\) is the speed of light (\(3 \times 10^8\) m/s).
02

Calculate the de Broglie wavelength of electrons

Now that we know the x-ray wavelength, we want electrons with the same wavelength. The de Broglie wavelength formula relates the wavelength of particles to its momentum, which is: \(\lambda_e = \frac{h}{p_e}\) where \(\lambda_e\) is the electron wavelength, and \(p_e\) is electron momentum. For the electrons to have the same wavelength as the x-rays, we need \(\lambda_e = \lambda_x\).
03

Find the electron momentum

Since we now know the electron wavelength, we can find the electron momentum. From the de Broglie wavelength formula: \(p_e = \frac{h}{\lambda_e}\) Now, substitute \(\lambda_x\) for \(\lambda_e\): \(p_e = \frac{h}{\lambda_x}\)
04

Relate electron momentum to its kinetic energy

We want to find the kinetic energy (\(K_e\)) of the electrons. The relationship between the electron momentum and kinetic energy is given by: \(K_e = \frac{p_e^2}{2m_e}\) where \(m_e\) is the mass of the electron (\(9.11 \times 10^{-31}\) kg).
05

Calculate the kinetic energy of electrons

Now, we can substitute the expression for electron momentum from Step 3 into the kinetic energy equation and solve for the kinetic energy of electrons. \(K_e = \frac{(h/\lambda_x)^2}{2m_e}\) We've previously calculated \(\lambda_x\) in Step 1, \(h\) and \(m_e\) are known constants, so we can now find \(K_e\). Once you have found the kinetic energy of electrons (\(K_e\)) in Joules, convert it to electron volts (eV) by dividing the value by \(1.6 \times 10^{-19}\). This will give you the kinetic energy of electrons required to produce the same diffraction pattern as the 16-keV x-rays using the same crystal.

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Most popular questions from this chapter

In the Davisson-Germer experiment (Section \(28.2),\) the electrons were accelerated through a \(54.0-\mathrm{V}\) potential difference before striking the target. (a) Find the de Broglie wavelength of the electrons. (b) Bragg plane spacings for nickel were known at the time; they had been determined through x-ray diffraction studies. The largest plane spacing (which gives the largest intensity diffraction maxima) in nickel is \(0.091 \mathrm{nm} .\) Using Bragg's law [Eq. ( \(25-15\) )], find the Bragg angle for the first-order maximum using the de Broglie wavelength of the electrons. (c) Does this agree with the observed maximum at a scattering angle of \(130^{\circ} ?\) [Hint: The scattering angle and the Bragg angle are not the same. Make a sketch to show the relationship between the two angles.]
A beam of neutrons is used to study molecular structure through a series of diffraction experiments. A beam of neutrons with a wide range of de Broglie wavelengths comes from the core of a nuclear reactor. In a time-offlight technique, used to select neutrons with a small range of de Broglie wavelengths, a pulse of neutrons is allowed to escape from the reactor by opening a shutter very briefly. At a distance of \(16.4 \mathrm{m}\) downstream, a second shutter is opened very briefly 13.0 ms after the first shutter. (a) What is the speed of the neutrons selected? (b) What is the de Broglie wavelength of the neutrons? (c) If each shutter is open for 0.45 ms, estimate the range of de Broglie wavelengths selected.
What is the ratio of the wavelength of a 0.100-keV photon to the wavelength of a 0.100-keV electron?
What are the possible values of \(L_{z}\) (the component of angular momentum along the \(z\) -axis) for the electron in the second excited state \((n=3)\) of the hydrogen atom?
A bullet leaves the barrel of a rifle with a speed of $300.0 \mathrm{m} / \mathrm{s} .\( The mass of the bullet is \)10.0 \mathrm{g} .$ (a) What is the de Broglie wavelength of the bullet? (b) Compare \(\lambda\) with the diameter of a proton (about \(1 \mathrm{fm}\) ). (c) Is it possible to observe wave properties of the bullet, such as diffraction? Explain.
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