An electron moving with speed v and a photon | vedclass

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(C) Given that the de Broglie wavelengths are equal: $\lambda_e = \lambda_{ph}$.Since $\lambda = \frac{h}{p}$,we have $p_e = p_{ph}$.The momentum of t

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$\frac{v}{2c}$

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(C) Given that the de Broglie wavelengths are equal: $\lambda_e = \lambda_{ph}$.Since $\lambda = \frac{h}{p}$,we have $p_e = p_{ph}$.The momentum of the electron is $p_e = \sqrt{2mK_e}$ and the momentum of the photon is $p_{ph} = \frac{E_{ph}}{c}$.Equating the momenta: $\sqrt{2mK_e} = \frac{E_{ph}}{c}$.Squaring both sides: $2mK_e = \frac{E_{ph}^2}{c^2}$.Rearranging for the ratio of kinetic energy of the electron $(K_e)$ to the energy of the photon $(E_{ph})$:$\frac{K_e}{E_{ph}} = \frac{E_{ph}}{2mc^2}$.Since $E_{ph} = p_{ph}c$ and $p_{ph} = p_e = mv$,we substitute $E_{ph} = (mv)c$.$\frac{K_e}{E_{ph}} = \frac{mvc}{2mc^2} = \frac{v}{2c}$.

An electron moving with speed $v$ and a photon moving with speed $c$ have the same de Broglie wavelength. The ratio of the kinetic energy of the electron to that of the photon is -

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