# Question #db111

May 4, 2017

4.82 keV photon

Energy $E = \frac{h c}{\lambda} \implies \lambda = \frac{6.63 \times {10}^{- 34} \cdot 3 \times {10}^{8}}{4.82 \times {10}^{3} \cdot \textcolor{red}{1.6 \times {10}^{- 19}}} \approx 0.26 \text{ nm}$

[Term in red: An eV is the energy required to increase the potential of an electron by 1 Volt so: $1 \text{eV" = e^(-) " J}$]

4.82 keV neutron

Here we do use the De Broglie relation $\lambda = \text{h/p}$.

The neutron's energy is: $E = {p}^{2} / \left(2 m\right) \implies p = \sqrt{2 E m} \implies \lambda = \frac{h}{\sqrt{2 E {m}_{n}}}$

That is:

$\frac{6.63 \times {10}^{- 34}}{\sqrt{2 \left(4.82 \times {10}^{3} \cdot \textcolor{red}{1.6 \times {10}^{- 19}}\right) 1.67 \times {10}^{- 27}}} \approx 0.4 \times {10}^{- 12} \text{ m}$