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(B) The de-Broglie wavelength $\lambda$ is related to the kinetic energy $K$ of an electron by the formula: $\lambda = \frac{h}{\sqrt{2mK}}$.Rearrangi

Vedclass · Accepted Answer

$16.8$

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(B) The de-Broglie wavelength $\lambda$ is related to the kinetic energy $K$ of an electron by the formula: $\lambda = \frac{h}{\sqrt{2mK}}$.Rearranging for $K$, we get: $K = \frac{h^2}{2m\lambda^2}$.Given $\lambda = 0.3 \, nm = 0.3 	imes 10^{-9} \, m$, $h = 6.626 	imes 10^{-34} \, J \cdot s$, and $m = 9.1 	imes 10^{-31} \, kg$.Substituting the values: $K = \frac{(6.626 	imes 10^{-34})^2}{2 	imes 9.1 	imes 10^{-31} 	imes (0.3 	imes 10^{-9})^2}$.$K = \frac{43.9 	imes 10^{-68}}{18.2 	imes 10^{-31} 	imes 0.09 	imes 10^{-18}} = \frac{43.9 	imes 10^{-68}}{1.638 	imes 10^{-48}} \approx 26.8 	imes 10^{-20} \, J$.To convert Joules to $eV$, divide by $1.6 	imes 10^{-19} \, J/eV$: $K \approx \frac{26.8 	imes 10^{-20}}{1.6 	imes 10^{-19}} \approx 16.75 \, eV \approx 16.8 \, eV$.

The kinetic energy of an electron with a de-Broglie wavelength of $0.3 \, nm$ is ............. $eV$.

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