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: Use the ready-made neutron formula.
Starting from \(\lambda=\dfrac{h}{\sqrt{2m_{n}E}}\) and putting in the fixed constants \(h\), \(m_{n}\) and \(1\,\text{eV}=1.6\times10^{-19}\,\text{J}\), the algebra collapses to a handy shortcut when the energy \(E\) is expressed in electron-volts:
\[ \lambda=\frac{0.286}{\sqrt{E(\text{in eV})}}\ \text{\AA} \]
This lets us skip repeated joule conversions.

Step 2: Put the energy in electron-volts.
\[ E=12.8\ \text{MeV}=12.8\times10^{6}\ \text{eV}=1.28\times10^{7}\ \text{eV} \]

Step 3: Take the square root of the energy.
\[ \sqrt{1.28\times10^{7}}=3.58\times10^{3} \]

Step 4: Evaluate the wavelength in angstrom.
\[ \lambda=\frac{0.286}{3.58\times10^{3}}=7.99\times10^{-5}\ \text{\AA} \]

Step 5: Convert to metres.
Since \(1\ \text{\AA}=10^{-10}\ \text{m}\),
\[ \lambda=7.99\times10^{-5}\times10^{-10}\ \text{m}=7.99\times10^{-15}\ \text{m} \]
which agrees with the direct calculation.

Result:
\[\boxed{\lambda\approx8.0\times10^{-15}\ \text{m}}\]