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(B) The de-Broglie wavelength of a particle is given by $\lambda = \frac{h}{p}$.For an electron with energy $E$ and mass $m$,the momentum $p_e = \sqrt

Vedclass · Accepted Answer

$\frac{1}{c}\left[\frac{E}{2m}\right]^{1/2}$

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(B) The de-Broglie wavelength of a particle is given by $\lambda = \frac{h}{p}$.For an electron with energy $E$ and mass $m$,the momentum $p_e = \sqrt{2mE}$. Thus,$\lambda_e = \frac{h}{\sqrt{2mE}}$.For a photon with energy $E$,the momentum $p_p = \frac{E}{c}$. Thus,$\lambda_p = \frac{h}{p_p} = \frac{hc}{E}$.The ratio of the wavelengths is $\frac{\lambda_e}{\lambda_p} = \frac{h}{\sqrt{2mE}} 	imes \frac{E}{hc} = \frac{1}{c} \frac{E}{\sqrt{2mE}} = \frac{1}{c} \sqrt{\frac{E}{2m}} = \frac{1}{c} \left[\frac{E}{2m}ight]^{1/2}$.

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is ($c$ = velocity of light in air).

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