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(a) What is the momentum of a 0.0100-nm-wavelength photon that could detect details of an atom? (b) What is its energy in MeV?
a) $6.63 \times 10^{-23} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$b) 0.124 $\mathrm{MeV}$
Physics 103
Chapter 29
Introduction to Quantum Physics
Quantum Physics
Simon Fraser University
Hope College
University of Sheffield
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and this problem were asked to calculate the mo mentum of a 0.1 nana meter wavelength. So mo mentum is our planks constant, divided by wavelength. So planks constant is 6.63 times 10 to the negative 34 jewels Time seconds divided by our wavelength in meters. So that 0.1 times 10 to the negative nine meters gives us a momenta MB of 6.63 times 10 to the negative. 23rd negative 23rd kilograms meters per second for pardon me for part B, whereas too determined this amount of energy in mega electron volts. So our energy is planks, constant times frequency. And since we have wavelength freaking convert frequency too speed of light divided by wavelength. So 6.63 times 10 to the negative. 23rd is planks constant. The speed of light is 2.99 times 10 to the eighth meters per second. Divided by our wavelength which is 0.0 one times 10 to the negative nine gives us an energy of zero point. Gives us an energy of 1.98 times 10 to the negative 14 jewels, but it wants it in electron volts so we can divide my jewels. But 1.602 times 10 to the negative 13 jewels, which is one mega electron volt to get us an energy of 0.124 mega electron volts.
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