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(A) The de-Broglie wavelength of an electron is given by $\lambda_e = \frac{h}{\sqrt{2 m_e K_e}}$,where $K_e$ is the kinetic energy of the electron.Sq

Vedclass · Accepted Answer

$1 : 100$

Vedclass · Answer

(A) The de-Broglie wavelength of an electron is given by $\lambda_e = \frac{h}{\sqrt{2 m_e K_e}}$,where $K_e$ is the kinetic energy of the electron.Squaring both sides,we get $\lambda_e^2 = \frac{h^2}{2 m_e K_e}$,which implies $K_e = \frac{h^2}{2 m_e \lambda_e^2}$.The energy of a photon is given by $K_p = \frac{hc}{\lambda_p}$.Given that $\lambda_e = \lambda_p = \lambda = 1.2 \ 	ext{Å} = 1.2 	imes 10^{-10} \ 	ext{m}$.The ratio of their energies is $\frac{K_e}{K_p} = \frac{h^2 / (2 m_e \lambda^2)}{hc / \lambda} = \frac{h}{2 m_e c \lambda}$.Substituting the values: $h = 6.63 	imes 10^{-34} \ 	ext{J} \cdot 	ext{s}$,$m_e = 9.11 	imes 10^{-31} \ 	ext{kg}$,$c = 3 	imes 10^8 \ 	ext{m/s}$,and $\lambda = 1.2 	imes 10^{-10} \ 	ext{m}$.$\frac{K_e}{K_p} = \frac{6.63 	imes 10^{-34}}{2 	imes 9.11 	imes 10^{-31} 	imes 3 	imes 10^8 	imes 1.2 	imes 10^{-10}} \approx \frac{6.63 	imes 10^{-34}}{6.56 	imes 10^{-31}} \approx 0.01 = \frac{1}{100}$.Thus,the ratio $K_e : K_p$ is $1 : 100$.

Both an electron and a photon have the same de-Broglie wavelength of $1.2 \ \text{Å}$. The ratio of their energies is nearly

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