Question

Consider a neutron (mass $m$) of kinetic energy $E$ and a photon of the same energy. Let $\lambda_n$ and $\lambda_p$ be the de Broglie wavelength of the neutron and the wavelength of the photon, respectively. Obtain an expression for $\frac{\lambda_n{\lambda_p}$.}

Hint:

For de Broglie wavelength problems, use $\lambda = \frac{h}{p}$ and carefully relate the momentum or energy for each particle or photon.

Answer 1

The equation for energy \(E\) is given as \(E = \frac{hc}{\lambda_p}\), which can be rearranged to \(\lambda_p = \frac{hc}{E}\). Using the relationship \(\lambda_n = \frac{h}{p}\), where \(p\) represents momentum, we substitute the expression for \(p\): \[\lambda_n = \frac{h}{p} = \frac{h}{\sqrt{2mE}}\] Substituting the expression for \(p\) again yields: \[\frac{\lambda_n}{\lambda_p} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{h}{Ehc}} = \frac{E}{\sqrt{2mc^2}}\] This results in the expression: \[\frac{\lambda_n}{\lambda_p} = \sqrt{\frac{E}{2mc^2}}\] This provides the required expression for \(\frac{\lambda_n}{\lambda_p}\).