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(C) The de-Broglie wavelength $\lambda$ is related to momentum $P$ by the equation $\lambda = \frac{h}{P}$.Since momentum $P = \sqrt{2m(K.E.)}$, we ca

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$1 	imes 10^{-21} \,J$

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(C) The de-Broglie wavelength $\lambda$ is related to momentum $P$ by the equation $\lambda = \frac{h}{P}$.Since momentum $P = \sqrt{2m(K.E.)}$, we can write $\lambda = \frac{h}{\sqrt{2m(K.E.)}}$.Squaring both sides and rearranging for kinetic energy $(K.E.)$, we get $K.E. = \frac{h^2}{2m\lambda^2}$.Given $h = 6.6 	imes 10^{-34} \,J \cdot s$, $m = 2 	imes 10^{-27} \,kg$, and $\lambda = 3.3 	imes 10^{-10} \,m$.Substituting these values:$K.E. = \frac{(6.6 	imes 10^{-34})^2}{2 	imes (2 	imes 10^{-27}) 	imes (3.3 	imes 10^{-10})^2}$$K.E. = \frac{43.56 	imes 10^{-68}}{4 	imes 10^{-27} 	imes 10.89 	imes 10^{-20}}$$K.E. = \frac{43.56 	imes 10^{-68}}{43.56 	imes 10^{-47}}$$K.E. = 1 	imes 10^{-21} \,J$.

$A$ particle of mass $2 \times 10^{-27} \,kg$ has a de-Broglie wavelength of $3.3 \times 10^{-10} \,m$. The kinetic energy of this particle is (Planck's constant $h = 6.6 \times 10^{-34} \,J \cdot s$).

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