The mass of a proton is 1.67 times 10^{-27} | vedclass

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(A) According to the de Broglie equation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.Given:$h = 6.626 	imes 10^{-34} \ J \cdot s$$m

Vedclass · Accepted Answer

$0.40$

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(A) According to the de Broglie equation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.Given:$h = 6.626 	imes 10^{-34} \ J \cdot s$$m = 1.67 	imes 10^{-27} \ kg$$v = 10^3 \ m/s$Substituting the values:$\lambda = \frac{6.626 	imes 10^{-34}}{1.67 	imes 10^{-27} 	imes 10^3}$$\lambda = \frac{6.626 	imes 10^{-34}}{1.67 	imes 10^{-24}}$$\lambda \approx 3.967 	imes 10^{-10} \ m$To convert meters to nanometers $(nm)$,multiply by $10^9$:$\lambda = 3.967 	imes 10^{-10} 	imes 10^9 \ nm = 0.3967 \ nm$Rounding off,we get $\lambda \approx 0.40 \ nm$.

The mass of a proton is $1.67 \times 10^{-27} \ kg$. If it is moving with a velocity of $10^3 \ m/s$,calculate its wavelength in nanometers.

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