Question

The de Broglie wavelengths of a photon and a proton are \(\lambda_1\) and \(\lambda_2\) respectively. Both have same energy \(E\). If mass of proton is \(m\) and speed of light is \(C\), then prove that \(\dfrac{\lambda_1}{\lambda_2} = C\sqrt{\dfrac{2m}{E}}\).

Hint:

Use \(\lambda = h/p\). For the photon \(p = E/C\); for the proton \(p = \sqrt{2mE}\). Then divide.

Answer 1

Step 1 (Common tool): Both wavelengths come from \(\lambda = h/p\). The whole proof is really about expressing each momentum \(p\) in terms of the shared energy \(E\).

Step 2 (Momentum of the photon): Light carries momentum \(p_1 = E/C\) because a photon obeys \(E = p_1 C\). So its wavelength is \(\lambda_1 = h/p_1 = hC/E\).

Step 3 (Momentum of the proton): A massive particle with kinetic energy \(E\) satisfies \(p_2 = \sqrt{2mE}\), obtained from \(E = \tfrac{1}{2}m v^2\) combined with \(p_2 = m v\). So \(\lambda_2 = h/\sqrt{2mE}\).

Step 4 (Divide by cancelling \(h\)): Instead of substituting fractions, divide the wavelengths directly:
\[ \frac{\lambda_1}{\lambda_2} = \frac{p_2}{p_1} = \frac{\sqrt{2mE}}{E/C} = \frac{C\sqrt{2mE}}{E} \]
Notice the ratio of wavelengths equals the reversed ratio of momenta, which is a neat cross-check.

Step 5 (Reduce the surd): Since \(\dfrac{\sqrt{2mE}}{E} = \sqrt{\dfrac{2m}{E}}\),
\[ \frac{\lambda_1}{\lambda_2} = C\sqrt{\frac{2m}{E}} \]
which is exactly what had to be shown.
\[\boxed{\dfrac{\lambda_1}{\lambda_2} = C\sqrt{\dfrac{2m}{E}}}\]