A proton of mass ' $m_P$ ' has same energy as that of a photon of wavelength ' $\lambda$ '. If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.

<p>To find the ratio of the de Broglie wavelength of a proton to the wavelength of a photon when they have the same energy, we follow these steps:</p> <p><p><strong>Energy Equivalence</strong>: </p> <p>Both the proton and the photon have the same energy, denoted by <code>E</code>.</p></p> <p><p><strong>Energy of the Photon</strong>: </p> <p>The energy of a photon with wavelength <code>λ</code> is given by: </p> <p>$ E = \frac{hc}{\lambda} $</p></p> <p><p><strong>Kinetic Energy of the Proton</strong>: </p> <p>Since the proton moves at a non-relativistic speed, its kinetic energy can be expressed as: </p> <p>$ \text{KE}_{\text{proton}} = \frac{1}{2}m_P v^2 = E $</p></p> <p><p><strong>Momentum of the Proton</strong>: </p> <p>The momentum <code>p</code> of the proton is derived from its kinetic energy: </p> <p>$ p = \sqrt{2m_P E} $</p></p> <p><p><strong>de Broglie Wavelength of the Proton</strong>: </p> <p>The de Broglie wavelength <code>λ_{proton}</code> is: </p> <p>$ \lambda_{\text{proton}} = \frac{h}{p} = \frac{h}{\sqrt{2m_P E}} $</p></p> <p><p><strong>Ratio of Wavelengths</strong>: </p> <p>The ratio of the de Broglie wavelength of the proton to the wavelength of the photon is: </p> <p>$ \frac{\lambda_{\text{proton}}}{\lambda_{\text{photon}}} = \left( \frac{h}{\sqrt{2m_P E}} \right) \times \left( \frac{E}{hc} \right) $ </p> <p>Simplifying, we get: </p> <p>$ \frac{\lambda_{\text{proton}}}{\lambda_{\text{photon}}} = \frac{1}{c} \sqrt{\frac{E}{2m_P}} $</p></p> <p>Thus, the ratio of the de Broglie wavelength of the proton to the wavelength of the photon is $\frac{1}{c} \sqrt{\frac{E}{2m_P}}$.</p>