Question

Find the de Broglie wavelength associated with a thermal neutron of mass \(m\) at absolute temperature \(T\). \[ k=\text{Boltzmann constant}, \qquad h=\text{Planck's constant} \]

• A. \[ \frac{h}{\sqrt{3mkT}} \]
• B. \[ \frac{h}{\sqrt{2mkT}} \]
• C. \[ \frac{h}{3mkT} \]
• D. \[ \frac{h}{\sqrt{mkT}} \]

Hint:

For a thermal particle, \[ \frac{p^2}{2m} = \frac32 kT. \] Then use \[ \lambda=\frac{h}{p}. \]

Answer 1

The Correct Option is A

Step 1: Link thermal energy to kinetic energy.
A thermal neutron shares the average thermal energy, so its kinetic energy is $\frac{3}{2}kT$, where $k$ is Boltzmann's constant.
Step 2: Write kinetic energy using momentum.
Kinetic energy can also be written as $K = \frac{p^2}{2m}$, which lets us pull out the momentum $p$.
Step 3: Set the two expressions equal.
\[ \frac{p^2}{2m} = \frac{3}{2}kT \]
Step 4: Solve for momentum.
Multiply both sides by $2m$: $p^2 = 3mkT$, so $p = \sqrt{3mkT}$.
Step 5: Apply the de Broglie relation.
The de Broglie wavelength is $\lambda = \frac{h}{p}$, where $h$ is Planck's constant.
Step 6: Substitute the momentum.
\[ \lambda = \frac{h}{\sqrt{3mkT}} \]
\[ \boxed{\lambda = \frac{h}{\sqrt{3mkT}}} \]