We are determined to provide the latest solutions related to all subjects FREE of charge!
Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.
Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program
Advertise your business, and reach millions of students around the world.
All the data tables that you may search for.
For Arabic Users, find a teacher/tutor in your City or country in the Middle East.
Find the Source, Textbook, Solution Manual that you are looking for in 1 click.
Need Help? We got you covered.
Neutrons are used to study structures of solids and their properties. What energy (and temperature) neutrons are needed if the atomic structures are of the size 0.060 nm?
Strategy We use the de Broglie wavelength relation to match the wavelength needed with an appropriate momentum. From the momentum we determine the kinetic energy and temperature from K=\frac{3}{2} k T.
We determine the neutron momentum from the de Broglie wavelength relation with \lambda=0.060 nm.
p=\frac{h}{\lambda}=\frac{6.63 \times 10^{-34} J \cdot s }{0.060 \times 10^{-9} m }=1.1 \times 10^{-23} kg \cdot m / s
Because we expect the kinetic energy to be low, we use a nonrelativistic relation to determine the kinetic energy.
\begin{aligned}K &=\frac{p^{2}}{2 m}=\frac{\left(1.1 \times 10^{-23} kg \cdot m / s \right)^{2}}{2\left(1.67 \times 10^{-27} kg \right)}=3.6 \times 10^{-20} J \\&=0.23 eV\end{aligned}
Thus we see that the nonrelativistic relation is certainly adequate here. In thermal equilibrium, the temperature is found from K=\frac{3}{2} k T.
T=\frac{2 K}{3 k}=\frac{2\left(3.6 \times 10^{-20} J \right)}{3\left(1.38 \times 10^{-23} J / K \right)}=1740 K
Such energy is easily obtained by thermalizing neutrons from a nuclear reactor.