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(C) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron.We know that kinetic

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$0.38$

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(C) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron.We know that kinetic energy $K = \frac{p^2}{2m}$,so $p = \sqrt{2mK}$.Substituting this into the wavelength formula: $\lambda = \frac{h}{\sqrt{2mK}}$.Rearranging for $K$: $K = \frac{h^2}{2m\lambda^2}$.Given values: $h = 6.6 	imes 10^{-34} \ J \ s$,$m = 9 	imes 10^{-31} \ kg$,and $\lambda = 2 \ nm = 2 	imes 10^{-9} \ m$.$K = \frac{(6.6 	imes 10^{-34})^2}{2 	imes (9 	imes 10^{-31}) 	imes (2 	imes 10^{-9})^2}$.$K = \frac{43.56 	imes 10^{-68}}{18 	imes 10^{-31} 	imes 4 	imes 10^{-18}} = \frac{43.56 	imes 10^{-68}}{72 	imes 10^{-49}} \approx 0.605 	imes 10^{-19} \ J$.To convert Joules to electron-volts $(eV)$,divide by $1.6 	imes 10^{-19} \ J/eV$:$K = \frac{0.605 	imes 10^{-19}}{1.6 	imes 10^{-19}} \approx 0.378 \ eV$.Rounding to the nearest option,we get $0.38 \ eV$.

If the de Broglie wavelength of an electron is $2 \ nm$,then its kinetic energy is nearly (Planck's constant $= 6.6 \times 10^{-34} \ J \ s$ and mass of electron $= 9 \times 10^{-31} \ kg$) (in $eV$)

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