If the mass of a neutron is 1.7 times 10^{-27 | vedclass

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(B) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum.Since kinetic energy $E = \frac{p^2}

Vedclass · Accepted Answer

$1.65 	imes 10^{-11} \; m$

Vedclass · Answer

(B) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum.Since kinetic energy $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.Thus,$\lambda = \frac{h}{\sqrt{2mE}}$.Given: $m = 1.7 	imes 10^{-27} \; kg$,$E = 3 \; eV = 3 	imes 1.6 	imes 10^{-19} \; J$,and $h = 6.6 	imes 10^{-34} \; J \cdot s$.Substituting the values:$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{2 	imes 1.7 	imes 10^{-27} 	imes 3 	imes 1.6 	imes 10^{-19}}}$$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{16.32 	imes 10^{-46}}}$$\lambda = \frac{6.6 	imes 10^{-34}}{4.04 	imes 10^{-23}}$$\lambda \approx 1.633 	imes 10^{-11} \; m$.Rounding to the nearest option,we get $\lambda = 1.65 	imes 10^{-11} \; m$.

If the mass of a neutron is $1.7 \times 10^{-27} \; kg$,then the de-Broglie wavelength of a neutron with energy $3 \; eV$ is (given $h = 6.6 \times 10^{-34} \; J \cdot s$):

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