What is the kinetic energy of a neutron assoc | vedclass

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.From this,the momentum $p$ is $p = \frac{h}{\lambda}$.The kinetic ene

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$6.69 	imes 10^{-21} \ J$

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.From this,the momentum $p$ is $p = \frac{h}{\lambda}$.The kinetic energy $K$ is related to momentum by $K = \frac{p^2}{2m}$.Substituting $p = \frac{h}{\lambda}$,we get $K = \frac{h^2}{2m\lambda^2}$.Given values: $h = 6.6 	imes 10^{-34} \ J \cdot s$,$m = 1.675 	imes 10^{-27} \ kg$,and $\lambda = 1.40 	imes 10^{-10} \ m$.$K = \frac{(6.6 	imes 10^{-34})^2}{2 	imes (1.675 	imes 10^{-27}) 	imes (1.40 	imes 10^{-10})^2}$.$K = \frac{43.56 	imes 10^{-68}}{2 	imes 1.675 	imes 10^{-27} 	imes 1.96 	imes 10^{-20}}$.$K = \frac{43.56 	imes 10^{-68}}{6.566 	imes 10^{-47}}$.$K \approx 6.634 	imes 10^{-21} \ J$.Rounding to the nearest option,we get $K \approx 6.69 	imes 10^{-21} \ J$.

What is the kinetic energy of a neutron associated with a de Broglie wavelength of $1.40 \times 10^{-10} \ m$? (Mass of neutron $m = 1.675 \times 10^{-27} \ kg$,$h = 6.6 \times 10^{-34} \ J \cdot s$).

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