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(C) The uncertainty in momentum is given by $\Delta p = m 	imes \Delta v$. Given: $\Delta p = 2 	imes 10^{-17} \ g \ cm \ s^{-1}$ and mass of electr

Vedclass · Accepted Answer

$2.1 	imes 10^{10} \ cm \ s^{-1}$

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(C) The uncertainty in momentum is given by $\Delta p = m 	imes \Delta v$. Given: $\Delta p = 2 	imes 10^{-17} \ g \ cm \ s^{-1}$ and mass of electron $m = 9.1 	imes 10^{-28} \ g$. We know that $\Delta v = \frac{\Delta p}{m}$. Substituting the values: $\Delta v = \frac{2 	imes 10^{-17} \ g \ cm \ s^{-1}}{9.1 	imes 10^{-28} \ g} \approx 0.2197 	imes 10^{11} \ cm \ s^{-1}$. This simplifies to $\Delta v \approx 2.197 	imes 10^{10} \ cm \ s^{-1}$,which is approximately $2.1 	imes 10^{10} \ cm \ s^{-1}$.

If the uncertainty in the position of an electron is associated with an uncertainty in momentum of $2 \times 10^{-17} \ g \ cm \ s^{-1}$,what will be the uncertainty in the velocity of the electron?

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