If the kinetic energy of an electron of mass | vedclass

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(C) Given,\ Kinetic energy $(KE) = 2.0 	imes 10^{-25} \ J$ \ Mass of electron $(m) = 9.0 	imes 10^{-31} \ kg$ \ Planck's constant $(h) = 6.63 	i

Vedclass · Accepted Answer

$1104$

Vedclass · Answer

(C) Given,\ Kinetic energy $(KE) = 2.0 	imes 10^{-25} \ J$ \ Mass of electron $(m) = 9.0 	imes 10^{-31} \ kg$ \ Planck's constant $(h) = 6.63 	imes 10^{-34} \ Js$ \ Using the de Broglie wavelength formula: $\lambda = \frac{h}{\sqrt{2m(KE)}}$ \ Substituting the values: $\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 9.0 	imes 10^{-31} 	imes 2.0 	imes 10^{-25}}}$ \ $\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{36 	imes 10^{-56}}} = \frac{6.63 	imes 10^{-34}}{6.0 	imes 10^{-28}}$ \ $\lambda = 1.105 	imes 10^{-6} \ m$ \ Converting to $nm$: $\lambda = 1.105 	imes 10^{-6} 	imes 10^9 \ nm = 1105 \ nm$ \ The closest approximate value is $1104.3 \ nm$.

If the kinetic energy of an electron of mass $9.0 \times 10^{-31} \ kg$ is $2.0 \times 10^{-25} \ J$,the wavelength of the electron in $nm$ is approximately (in $.3$)

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