Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]

The de Broglie wavelength for a particle of mass m and kinetic energy E is given by:

$$\lambda_{\text{n}} = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$$

The wavelength of a photon with energy E is given by:

$$\lambda_{\text{p}} = \frac{hc}{E}$$

Where h is Planck's constant and c is the speed of light.

Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]

To find the ratio \(\frac{\lambda_{\text{n}}}{\lambda_{\text{p}}}\), we substitute the expressions from Step 1:

$$\frac{\lambda_{\text{n}}}{\lambda_{\text{p}}} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}}$$

Rearranging the terms for simplification:

$$\frac{\lambda_{\text{n}}}{\lambda_{\text{p}}} = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc}$$

Canceling h and simplifying \(\frac{E}{\sqrt{E}}\) to \(\sqrt{E}\):

$$\frac{\lambda_{\text{n}}}{\lambda_{\text{p}}} = \frac{1}{\sqrt{2mE}} \times \frac{E}{c} = \frac{\sqrt{E}}{c\sqrt{2m}}$$

Step 3 (Final Concluded Answer): [0.5 Mark]

The final expression for the ratio of the wavelengths is:

$$\frac{\lambda_{\text{n}}}{\lambda_{\text{p}}} = \frac{1}{c} \sqrt{\frac{E}{2m}}$$

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💡 EL10 Concept Insight:

Think of it this way: a photon is pure energy moving at the speed limit of the universe (c), while a neutron is a heavy "ball" of matter; because the neutron has mass, it needs to move much slower than a photon to have the same energy, which makes its de Broglie wavelength behave differently than the photon's.