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(A) Given that the de Broglie wavelength of the electron $\lambda_e$ is equal to the wavelength of the photon $\lambda_p$,so $\lambda_e = \lambda_p =

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$1/4$

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(A) Given that the de Broglie wavelength of the electron $\lambda_e$ is equal to the wavelength of the photon $\lambda_p$,so $\lambda_e = \lambda_p = \lambda$.The kinetic energy of the electron is $K_e = \frac{1}{2} m_e v_e^2$.The energy of the photon is $E_p = \frac{hc}{\lambda_p}$.From the de Broglie relation,$\lambda_e = \frac{h}{m_e v_e}$,which implies $m_e = \frac{h}{\lambda_e v_e}$.Substituting $m_e$ into the kinetic energy expression: $K_e = \frac{1}{2} \left( \frac{h}{\lambda_e v_e} ight) v_e^2 = \frac{h v_e}{2 \lambda_e}$.Now,the ratio of the kinetic energy of the electron to the energy of the photon is:$\frac{K_e}{E_p} = \frac{h v_e / 2 \lambda_e}{hc / \lambda_p} = \frac{v_e}{2c}$.Given $v_e = 1.5 	imes 10^8 \, m/s$ and $c = 3 	imes 10^8 \, m/s$:$\frac{K_e}{E_p} = \frac{1.5 	imes 10^8}{2 	imes 3 	imes 10^8} = \frac{1.5}{6} = \frac{1}{4}$.

The de Broglie wavelength of an electron moving with a velocity $1.5 \times 10^8 \, m/s$ is equal to that of a photon. The ratio of the kinetic energy of the electron to the energy of the photon is

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