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(D) The de-Broglie wavelength of an electron is given by $\lambda_e = \frac{h}{mv}$,where $v$ is the velocity of the electron.The wavelength of a phot

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$1/4$

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(D) The de-Broglie wavelength of an electron is given by $\lambda_e = \frac{h}{mv}$,where $v$ is the velocity of the electron.The wavelength of a photon is given by $\lambda_p = \frac{h}{p_p} = \frac{hc}{E_p}$,where $E_p$ is the energy of the photon.Given that $\lambda_e = \lambda_p$,we have $\frac{h}{mv} = \frac{hc}{E_p}$.This implies $E_p = mvc$.The kinetic energy of the electron is $K_e = \frac{1}{2}mv^2$.The ratio of the kinetic energy of the electron to that of the photon is $\frac{K_e}{E_p} = \frac{\frac{1}{2}mv^2}{mvc} = \frac{v}{2c}$.Substituting the given values $v = 1.5 	imes 10^8 \ m/s$ 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 of $1.5 \times 10^8 \ m/s$ is equal to that of a photon. What is the ratio of the kinetic energy of the electron to that of the photon? (Given: $c = 3 \times 10^8 \ m/s$)

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