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(B) Given that the energy of the photon $E$ is equal to the kinetic energy of the proton $K_p = E$.For a proton,the de-Broglie wavelength $\lambda_1$

Vedclass · Accepted Answer

$E^{1/2}$

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(B) Given that the energy of the photon $E$ is equal to the kinetic energy of the proton $K_p = E$.For a proton,the de-Broglie wavelength $\lambda_1$ is given by $\lambda_1 = \frac{h}{p}$,where $p$ is the momentum of the proton.Since $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.Thus,$\lambda_1 = \frac{h}{\sqrt{2mE}}$.For a photon,the wavelength $\lambda_2$ is given by $\lambda_2 = \frac{hc}{E}$.Now,calculating the ratio $\frac{\lambda_1}{\lambda_2}$:$\frac{\lambda_1}{\lambda_2} = \frac{h / \sqrt{2mE}}{hc / E} = \frac{h}{\sqrt{2mE}} \cdot \frac{E}{hc} = \frac{E}{c \sqrt{2mE}} = \frac{\sqrt{E}}{c \sqrt{2m}}$.Since $c$ and $m$ are constants,$\frac{\lambda_1}{\lambda_2} \propto \sqrt{E}$ 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 the proton,$\lambda_2$ is the wavelength associated with the photon,and the energy of the photon is $E$,then $(\lambda_1 / \lambda_2)$ is proportional to

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