The wavelength of an electron is 10^3 nm. Wh | vedclass

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(C) According to the de-Broglie relation, the wavelength $\lambda$ is related to momentum $p$ by the formula: $\lambda = \frac{h}{p}$.Given values: $h

Vedclass · Accepted Answer

$6.625 	imes 10^{-28}$

Vedclass · Answer

(C) According to the de-Broglie relation, the wavelength $\lambda$ is related to momentum $p$ by the formula: $\lambda = \frac{h}{p}$.Given values: $h = 6.625 	imes 10^{-34} \ J \ s$ and $\lambda = 10^3 \ nm = 10^3 	imes 10^{-9} \ m = 10^{-6} \ m$.Rearranging the formula for momentum: $p = \frac{h}{\lambda}$.Substituting the values: $p = \frac{6.625 	imes 10^{-34} \ J \ s}{10^{-6} \ m} = 6.625 	imes 10^{-28} \ kg \ m \ s^{-1}$.

The wavelength of an electron is $10^3 \ nm$. What is its momentum (in $kg \ m \ s^{-1}$)? $(h = 6.625 \times 10^{-34} \ J \ s)$

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