Question

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

(Mass of an electron is $9\times {10}^{-28}\text{g})$

• A. $1\times {10}^{11}{\text{cm s}}^{-1}$
• B. $1\times {10}^9{\text{cm s}}^{-1}$
• C. $1\times {10}^6{\text{cm s}}^{-1}$
• D. $1\times {10}^5{\text{cm s}}^{-1}$

Answer 1

The correct option is B $1\times {10}^9{\text{cm s}}^{-1}$

Step 1: Given uncertainty in momentum of the electron,
$\left(\Delta p\right)=1\times {10}^{-21}{\text{kg cm s}}^{-1}$.
Mass of an electron,
$\left(m\right)=9.1\times {10}^{-31}\text{kg}$
To find the minimum uncertainty in the velocity of the electron,
We know that, $\Delta p=m\Delta v$
$\Delta v=\frac{\Delta p}{m}$
$\Delta v=\frac{1\times {10}^{-21}}{9.1\times {10}^{-31}}$

$\Delta v=1.09\times {10}^9{\text{cm s}}^{-1}$