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(A) The de-Broglie wavelength of an electron is given by $\lambda_e = \frac{h}{m_e v_e}$,where $v_e = c/2$. So,$\lambda_e = \frac{h}{m_e (c/2)} = \fra

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$1: 4$

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(A) The de-Broglie wavelength of an electron is given by $\lambda_e = \frac{h}{m_e v_e}$,where $v_e = c/2$. So,$\lambda_e = \frac{h}{m_e (c/2)} = \frac{2h}{m_e c}$.For a photon,the wavelength is $\lambda_p = \frac{h}{p_p} = \frac{h}{E_p/c} = \frac{hc}{E_p}$,where $E_p$ is the energy of the photon.Given $\lambda_e = \lambda_p$,we have $\frac{2h}{m_e c} = \frac{hc}{E_p}$.This implies $E_p = \frac{m_e c^2}{2}$.The kinetic energy of the electron is $K_e = \frac{1}{2} m_e v_e^2 = \frac{1}{2} m_e (c/2)^2 = \frac{1}{8} m_e c^2$.The ratio of kinetic energies is $\frac{K_e}{E_p} = \frac{\frac{1}{8} m_e c^2}{\frac{1}{2} m_e c^2} = \frac{1}{4}$.

The de-Broglie wavelength of an electron moving with a velocity $v = c / 2$ ($c$ is the velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of the electron and the photon is:

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