Solutions/Answers to Exercises
Exercise 1 What is the wavelength, in meters, of an electron (mass = 9.11 x 10-31 kg) that is accelerated to 4.5% the speed of light, c?
First, determine the speed, ν:
ν = 0.045 x (2.998 x 108 m/s) = 1.35 x 107 m/s
1 J = kg⋅m2/s2
\(\displaystyle \lambda\;=\;\frac{6.626\times\;10^{-34}\;\frac{kg⋅m^2}{s^2}⋅s}{9.11\times\;10^{-31}kg\times\;(1.35\times\;10^7\;m/s)}\;=\;\mathbf{5.39\times\;10^{-11}\;m}\)
Exercise 2 What is the wavelength of a butterfly with a mass of 0.500 g and a speed of 6.0 m/s? Can we observe this wavelength?
The mass of the butterfly is 0.000500 kg
\(\displaystyle \lambda\;=\;\frac{6.626\times\;10^{-34}\;\frac{kg⋅m^2}{s^2}⋅s}{0.000500\;kg\times\;(6.0\;m/s)}\;=\;\mathbf{2.2\times\;10^{-31}\;m}\)
No, we cannot observe this small wavelength.
Exercise 3 What speed, ν, would an electron (mass = 9.11 x 10-28 g) be traveling to have a de Broglie wavelength of 625 nm?
9.11 x 10-28 g is equal to 9.11 x 10-31 kg and 625 nm = 6.25 x 10-7 m
Solve \(\displaystyle \lambda\;=\;\frac{h}{m\nu}\) for ν
\(\displaystyle \nu\;=\;\frac{h}{m\lambda}\;=\;\frac{6.626\times\;10^{-34}\;\frac{kg⋅m^2}{s^2}⋅s}{9.11\times\;10^{-31}kg\times\;(6.25\times\;10^{-7}\;m)}\;=\;\mathbf{1.16\times\;10^3\;m/s}\)