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(B) The energy of a photon is given by $E = \frac{hc}{\lambda_2}$,so the wavelength of the photon is $\lambda_2 = \frac{hc}{E}$.For a proton,the kinet

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$E^{1/2}$

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(B) The energy of a photon is given by $E = \frac{hc}{\lambda_2}$,so the wavelength of the photon is $\lambda_2 = \frac{hc}{E}$.For a proton,the kinetic energy is $E = \frac{p^2}{2m}$,which gives the momentum $p = \sqrt{2mE}$.The de-Broglie wavelength of the proton is $\lambda_1 = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.Now,calculating the ratio $\frac{\lambda_1}{\lambda_2}$:$\frac{\lambda_1}{\lambda_2} = \frac{h / \sqrt{2mE}}{hc / E} = \frac{h}{\sqrt{2mE}} \cdot \frac{E}{hc} = \frac{E}{\sqrt{2mE} \cdot c} = \frac{\sqrt{E}}{\sqrt{2m} \cdot c}$.Since $m$ and $c$ are constants,we have $\frac{\lambda_1}{\lambda_2} \propto \sqrt{E}$ or $E^{1/2}$.

The energy of a photon is equal to the kinetic energy of a proton. If $\lambda_1$ is the de-Broglie wavelength of a proton and $\lambda_2$ is the wavelength associated with the photon,and if the energy of the photon is $E$,then $(\lambda_1 / \lambda_2)$ is proportional to

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