Kinetic energy of a proton is equal to energy | vedclass

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(A) For a proton with kinetic energy $E$,the momentum $p$ is given by $E = \frac{p^2}{2m}$,where $m$ is the mass of the proton.Thus,$p = \sqrt{2mE}$.T

Vedclass · Accepted Answer

$\frac{1}{2}$

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(A) For a proton with kinetic energy $E$,the momentum $p$ is given by $E = \frac{p^2}{2m}$,where $m$ is the mass of the proton.Thus,$p = \sqrt{2mE}$.The de-Broglie wavelength of the proton is $\lambda_1 = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.For a photon with energy $E$,the wavelength $\lambda_2$ is given by $E = \frac{hc}{\lambda_2}$,which implies $\lambda_2 = \frac{hc}{E}$.Taking the ratio of the two wavelengths:$\frac{\lambda_1}{\lambda_2} = \frac{h}{\sqrt{2mE}} \cdot \frac{E}{hc} = \frac{1}{c} \sqrt{\frac{E}{2m}}$.Therefore,$\frac{\lambda_1}{\lambda_2} \propto E^{1/2}$.Comparing this with $\frac{\lambda_1}{\lambda_2} \propto E^n$,we get $n = \frac{1}{2}$.

Kinetic energy of a proton is equal to energy $E$ of a photon. Let $\lambda_1$ be the de-Broglie wavelength of the proton and $\lambda_2$ be the wavelength of the photon. If $\frac{\lambda_1}{\lambda_2} \propto E^{n}$,then the value of $n$ is

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