If E_p and E_e represent the kinetic energy o | vedclass

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(B) For a photon,the energy is $E_p = h
u = hc / \lambda_p$. For an electron,the de-Broglie wavelength is $\lambda_e = h / p_e$,so $p_e = h / \lambda

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$1 	imes 10^{-2}$

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(B) For a photon,the energy is $E_p = h
u = hc / \lambda_p$. For an electron,the de-Broglie wavelength is $\lambda_e = h / p_e$,so $p_e = h / \lambda_e$. The kinetic energy of the electron is $E_e = p_e^2 / (2m) = h^2 / (2m \lambda_e^2)$. Given $\lambda_p = 2\lambda_e$,we have $\lambda_e = \lambda_p / 2$. Substituting this into the electron energy expression: $E_e = h^2 / (2m (\lambda_p / 2)^2) = 2h^2 / (m \lambda_p^2)$. Now,find the ratio $E_e / E_p$: $E_e / E_p = [2h^2 / (m \lambda_p^2)] / [hc / \lambda_p] = 2h / (mc \lambda_p)$. Since $\lambda_e = h / (m v_e)$,we have $\lambda_p = 2 \lambda_e = 2h / (m v_e)$. Substitute $\lambda_p$ into the ratio: $E_e / E_p = 2h / (mc \cdot (2h / (m v_e))) = v_e / c$. Given $v_e = C / 100$,the ratio is $E_e / E_p = (C / 100) / C = 1 / 100 = 1 	imes 10^{-2}$.

If $E_p$ and $E_e$ represent the kinetic energy of a photon and an electron respectively. If the de-Broglie wavelength $\lambda_p$ of a photon is twice the de-Broglie wavelength $\lambda_e$ of an electron,then $E_e / E_p$ is (Speed of electron $= C/100$,where $C$ is the velocity of light).

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