The wavelength of a microscopic particle of m | vedclass

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(B) Given,mass of particle $(m) = 9.1 	imes 10^{-31} \ kg$. Wavelength $(\lambda) = 182 \ nm = 182 	imes 10^{-9} \ m$. According to the de-Broglie e

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$7.28 	imes 10^{-24}$

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(B) Given,mass of particle $(m) = 9.1 	imes 10^{-31} \ kg$. Wavelength $(\lambda) = 182 \ nm = 182 	imes 10^{-9} \ m$. According to the de-Broglie equation,$\lambda = \frac{h}{mv}$,so velocity $v = \frac{h}{m \lambda}$. Substituting the values: $v = \frac{6.625 	imes 10^{-34}}{9.1 	imes 10^{-31} 	imes 182 	imes 10^{-9}} \ m/s$. $v = \frac{6.625 	imes 10^{-34}}{1656.2 	imes 10^{-40}} \ m/s \approx 4000 \ m/s = 4 	imes 10^3 \ m/s$. Kinetic energy $(KE) = \frac{1}{2} mv^2 = \frac{1}{2} 	imes (9.1 	imes 10^{-31} \ kg) 	imes (4 	imes 10^3 \ m/s)^2$. $KE = 0.5 	imes 9.1 	imes 10^{-31} 	imes 16 	imes 10^6 \ J = 72.8 	imes 10^{-25} \ J = 7.28 	imes 10^{-24} \ J$.

The wavelength of a microscopic particle of mass $9.1 \times 10^{-31} \ kg$ is $182 \ nm$. Its kinetic energy in $J$ is: $(h = 6.625 \times 10^{-34} \ J \ s)$

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