Video Transcript
A diffraction pattern formed by X-rays of wavelength 0.170 nanometers determines that the spacing between crystalline planes in an NaCl crystal is 0.281 nanometers. A neutron beam produces diffraction intensity peaks at the same locations that such peaks occur in the X-ray diffraction pattern. What is the energy of neutrons in the beam?
We’re told in this statement that the X-rays incident on the crystal have a wavelength of 0.170 nanometers; we’ll call that value 𝜆. We’re also told that the spacing between layers of the salt crystal is 0.281 nanometers; we’ll call that value 𝑑.
Given that a neutron beam produces a diffraction intensity pattern with peaks at the same locations as the pattern produced by X-rays, we want to know the energy 𝐸 of the neutrons in the beam. In this example, our sodium chloride crystal with layers separated by a distance 𝑑 has X-rays of wavelength 𝜆 impinge on it.
Then we have a second beam, this time made of neutrons, impinge on the same crystal. That means the energy of the second beam must match the energy of the first beam.
So what is the energy of the photons in the first beam, the beam of X-rays? Photon energy 𝐸 is equal to Planck’s constant ℎ times frequency 𝑓, which is also equal to ℎ times 𝑐 over 𝜆. Applying this relationship to our scenario, 𝐸 equals ℎ𝑐 over 𝜆.
If we assume ℎ is exactly 6.626 times 10 to the negative 34th joule seconds and 𝑐 is exactly 3.00 times 10 to the eighth meters per second, then we can plug those values into this expression, being careful to write- being careful with our wavelength value to use units of meters.
There’s one last step before we calculate our answer for 𝐸. If we calculated 𝐸 now, we would get an answer in units of joules. It would be correct, but since we’re working with small particles, an answer in units of electron volts might be more helpful.
Since one electron volt is equal to 1.60218 times 10 to the negative 19th joules, if we divide our result by that conversion factor, we get an answer in electron volts. When we calculate that value, we find it to be 7.30 times 10 to the third electron volts, or 7.30 kilo electron volts. That’s the energy of the neutrons in the beam.
