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(C) The uncertainty in momentum is given by $\Delta p = m \Delta v$,where $m$ is the mass and $\Delta v$ is the uncertainty in velocity.Given $\Delta

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$1 	imes 10^9 \ cm \ s^{-1}$

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(C) The uncertainty in momentum is given by $\Delta p = m \Delta v$,where $m$ is the mass and $\Delta v$ is the uncertainty in velocity.Given $\Delta p = 1 	imes 10^{-18} \ g \ cm \ s^{-1}$ and $m = 9 	imes 10^{-28} \ g$.Substituting these values into the equation:$1 	imes 10^{-18} = (9 	imes 10^{-28}) 	imes \Delta v$$\Delta v = \frac{1 	imes 10^{-18}}{9 	imes 10^{-28}} \ cm \ s^{-1}$$\Delta v = 0.111 	imes 10^{10} \ cm \ s^{-1} \approx 1.1 	imes 10^9 \ cm \ s^{-1}$.Rounding to the nearest provided option,the value is $1 	imes 10^9 \ cm \ s^{-1}$.

The measurement of the electron position is associated with an uncertainty in momentum,which is equal to $1 \times 10^{-18} \ g \ cm \ s^{-1}$. The uncertainty in electron velocity is (mass of an electron is $9 \times 10^{-28} \ g$):

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