The energy of a photon is equal to the kineti | vedclass

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(B) For a photon,the energy $E$ is related to its wavelength $\lambda_2$ by the equation $E = \frac{hc}{\lambda_2}$.Thus,$\lambda_2 = \frac{hc}{E}$  -

Vedclass · Accepted Answer

$E^{1/2}$

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(B) For a photon,the energy $E$ is related to its wavelength $\lambda_2$ by the equation $E = \frac{hc}{\lambda_2}$.Thus,$\lambda_2 = \frac{hc}{E}$  --- $(i)$For a proton with kinetic energy $E$,the de Broglie wavelength $\lambda_1$ is given by $\lambda_1 = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$  --- $(ii)$Taking the ratio $\frac{\lambda_1}{\lambda_2}$:$\frac{\lambda_1}{\lambda_2} = \frac{h / \sqrt{2mE}}{hc / E} = \frac{h}{\sqrt{2mE}} 	imes \frac{E}{hc} = \frac{E^{1/2}}{c\sqrt{2m}}$Since $c$,$m$,and $h$ are constants,we have $\frac{\lambda_1}{\lambda_2} \propto E^{1/2}$.

The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is $E$. If $\lambda_1$ and $\lambda_2$ are the de Broglie wavelengths of the proton and the photon respectively,then find the ratio $\lambda_1 / \lambda_2$.

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