An electron has kinetic energy 2.8 times 10^{ | vedclass

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(C) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$.Since kinetic energy $KE = \frac{p^2}{2m}$,we have $p = \sqrt{

Vedclass · Accepted Answer

$9.28 	imes 10^{-8} \ m$

Vedclass · Answer

(C) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$.Since kinetic energy $KE = \frac{p^2}{2m}$,we have $p = \sqrt{2mKE}$.Substituting this into the wavelength formula: $\lambda = \frac{h}{\sqrt{2mKE}}$.Given $h = 6.62 	imes 10^{-34} \ J \cdot s$,$m = 9.1 	imes 10^{-31} \ kg$,and $KE = 2.8 	imes 10^{-23} \ J$.$\lambda = \frac{6.62 	imes 10^{-34}}{\sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 2.8 	imes 10^{-23}}}$.$\lambda = \frac{6.62 	imes 10^{-34}}{\sqrt{50.96 	imes 10^{-54}}} = \frac{6.62 	imes 10^{-34}}{7.138 	imes 10^{-27}} \approx 9.28 	imes 10^{-8} \ m$.

An electron has kinetic energy $2.8 \times 10^{-23} \ J$. The de-Broglie wavelength will be nearly $(m_e = 9.1 \times 10^{-31} \ kg)$.

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