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

Concept: The de Broglie wavelength of any particle is \(\lambda = h/p\), where \(p\) is its momentum. A photon and a proton relate energy to momentum differently, so we handle each separately and then take the ratio.

Step 1 (Photon wavelength \(\lambda_1\)): A photon of energy \(E\) has momentum \(p_{ph} = E/C\) (since for a photon \(E = pC\)). Therefore
\[ \lambda_1 = \frac{h}{p_{ph}} = \frac{h}{E/C} = \frac{hC}{E} \]

Step 2 (Proton wavelength \(\lambda_2\)): The proton is non-relativistic, so its kinetic energy \(E\) links to momentum by \(E = \dfrac{p^2}{2m}\). Hence \(p = \sqrt{2mE}\) and
\[ \lambda_2 = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \]

Step 3 (Take the ratio):
\[ \frac{\lambda_1}{\lambda_2} = \frac{hC/E}{\,h/\sqrt{2mE}\,} = \frac{hC}{E}\cdot\frac{\sqrt{2mE}}{h} = \frac{C\sqrt{2mE}}{E} \]

Step 4 (Simplify): Write \(\dfrac{\sqrt{2mE}}{E} = \sqrt{\dfrac{2mE}{E^2}} = \sqrt{\dfrac{2m}{E}}\). Thus
\[ \frac{\lambda_1}{\lambda_2} = C\sqrt{\frac{2m}{E}} \]

Conclusion: The required relation is proved.
\[\boxed{\dfrac{\lambda_1}{\lambda_2} = C\sqrt{\dfrac{2m}{E}}}\]