Kinetic energy of a proton is equal to energy | vedclass

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(D) For a proton,the kinetic energy $E$ is related to its momentum $p$ by the equation $E = \frac{p^2}{2m}$,where $m$ is the mass of the proton.From t

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$0.5$

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(D) For a proton,the kinetic energy $E$ is related to its momentum $p$ by the equation $E = \frac{p^2}{2m}$,where $m$ is the mass of the proton.From this,the momentum is $p = \sqrt{2mE}$.The de-Broglie wavelength of the proton is $\lambda_1 = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.For a photon,the energy $E$ is related to its wavelength $\lambda_2$ by the equation $E = \frac{hc}{\lambda_2}$,where $c$ is the speed of light.Thus,the wavelength of the photon is $\lambda_2 = \frac{hc}{E}$.Now,taking the ratio of the wavelengths:$\frac{\lambda_1}{\lambda_2} = \left( \frac{h}{\sqrt{2mE}} ight) 	imes \left( \frac{E}{hc} ight) = \frac{1}{c} \sqrt{\frac{E}{2m}}$.This simplifies to $\frac{\lambda_1}{\lambda_2} = \left( \frac{1}{c\sqrt{2m}} ight) E^{1/2}$.Comparing this with $\frac{\lambda_1}{\lambda_2} \propto E^n$,we find that $n = 1/2 = 0.5$.

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 $\left(\frac{\lambda_1}{\lambda_2}\right) \propto E^{n}$,then the value of $n$ is:

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