91Ó°ÊÓ

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.]

Step by step solution

01

Part (a): Find the de Broglie wavelength

To find the de Broglie wavelength of the electrons, we'll use the following equation: \(\lambda = \dfrac{h}{p}\), where \(h\) is Planck's constant (\(6.626 \times 10^{-34}\,\mathrm{Js}\)) and \(p\) is the momentum of the electron, which can be found by using the energy-momentum relationship: \(E=\dfrac{p^2}{2m}\). In our case, the energy of the electrons is given by \(E=eV\), where \(e\) is the charge of an electron (\(1.602 \times 10^{-19}\,\mathrm{C}\)) and \(V\) is the potential difference (54 V). Therefore, we have \(\lambda = \dfrac{h}{\sqrt{2m(eV)}}\).

02

Calculate the momentum of the electron

First, calculate the energy of the electrons \(E=eV\), then use the energy-momentum relationship \(E=\dfrac{p^2}{2m}\) to find the momentum \(p\). In this case, \(E =(1.602\times 10^{-19}\,\mathrm{C})(54\,\mathrm{V}) = 8.65 \times 10^{-18}\,\mathrm{J}\) and \(m = 9.11\times 10^{-31} \,\mathrm{kg}\) (mass of an electron). Therefore, we have \(p = \sqrt{2mE} = \sqrt{(2)(9.11\times 10^{-31}\,\mathrm{kg})(8.65\times 10^{-18}\,\mathrm{J})} = 1.393 \times 10^{-24}\,\mathrm{kg\,m/s}\).

03

Calculate the de Broglie wavelength of the electron

Using the momentum calculated in step 1, find the de Broglie wavelength of the electron using the equation \(\lambda = \dfrac{h}{p}\). In this case, \(\lambda = \dfrac{6.626\times 10^{-34}\,\mathrm{Js}}{1.393\times 10^{-24}\,\mathrm{kg\,m/s}} = 4.761 \times 10^{-11}\,\mathrm{m}\) or \(47.61\,\mathrm{pm}\).

04

Part (b): Determine the Bragg angle

To find the Bragg angle (\(\theta\)) for the first-order maximum, we'll use Bragg's law: \(2d\sin{\theta}=n\lambda\), where \(d\) is the plane spacing (0.091 nm), \(n\) is the order of the maximum (1 in our case), and \(\lambda\) is the de Broglie wavelength. We'll rearrange the equation to solve for \(\theta\): \(\theta = \arcsin{(\dfrac{n\lambda}{2d})}\).

05

Calculate the Bragg angle

Using the de Broglie wavelength and plane spacing given in the problem, calculate the Bragg angle \(\theta\). In this case, \(\theta = \arcsin{(\dfrac{(1)(47.61\,\mathrm{pm})}{(2)(91\,\mathrm{pm})})} = \arcsin{(0.262)} = 15.3^{\circ}\).

06

Part (c): Compare the Bragg angle with the observed angle

The Bragg angle \(\theta\) is related to the observed scattering angle (given as \(130^{\circ}\)). To find the relationship between the two angles, we can notice that \(\theta\) and the scattering angle add up to \(90^{\circ}\). In other words, if we draw a sketch of the experiment, we'll see that when beam reflects off the plane, the angle between the incident and reflected beams forms a right angle. Therefore, we have: \(Bragg \, angle\, \theta + Scattering \, angle = 90^{\circ}\).

07

Compare the Bragg angle with the observed angle

Plugging in the Bragg angle we found in step 3 into the above equation, we have: \((15.3^{\circ}) + Scattering \, angle = 90^{\circ}\). Solving for the scattering angle, we get \(Scattering \, angle = 90^{\circ} - 15.3^{\circ} = 74.7^{\circ}\). Since this value (\(74.7^{\circ}\)) does not agree with the observed maximum at a scattering angle of \(130^{\circ}\), the answer to part (c) is that it does not agree with the observed maximum.

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