Question

Calculate the de Broglie wavelength of a neutron of energy \(12{\cdot}8\ \text{MeV}\). (Given: mass of neutron \(m_{n}=1{\cdot}67\times10^{-27}\ \text{kg}\), \(h=6{\cdot}6\times10^{-34}\ \text{J s}\), \(1\ \text{eV}=1{\cdot}6\times10^{-19}\ \text{J}\).)

Hint:

Use \(\lambda=h/\sqrt{2mE}\); first convert 12.8 MeV into joules.

Answer 1

Step 1: Write the de Broglie relation in terms of kinetic energy.
The de Broglie wavelength is \(\lambda=\dfrac{h}{p}\), where \(p\) is the momentum. For a non-relativistic particle of kinetic energy \(E\) and mass \(m\), the momentum is
\[ p=\sqrt{2mE} \]
Therefore
\[ \lambda=\frac{h}{\sqrt{2mE}} \]

Step 2: Convert the energy to joules.
\[ E=12{\cdot}8\ \text{MeV}=12{\cdot}8\times10^{6}\times1{\cdot}6\times10^{-19}\ \text{J}=2{\cdot}048\times10^{-12}\ \text{J} \]

Step 3: Compute \(2mE\).
\[ 2mE=2\times\left(1{\cdot}67\times10^{-27}\right)\times\left(2{\cdot}048\times10^{-12}\right)=6{\cdot}84\times10^{-39} \]

Step 4: Take the square root to get the momentum.
\[ \sqrt{2mE}=\sqrt{6{\cdot}84\times10^{-39}}=8{\cdot}27\times10^{-20}\ \text{kg m s}^{-1} \]

Step 5: Divide to get the wavelength.
\[ \lambda=\frac{6{\cdot}6\times10^{-34}}{8{\cdot}27\times10^{-20}}=7{\cdot}98\times10^{-15}\ \text{m} \]

Result:
\[\boxed{\lambda\approx8{\cdot}0\times10^{-15}\ \text{m}\;(=8{\cdot}0\ \text{fm})}\]