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

Options

A. `1 xx 10^6 cm s^-1`
B. `1 xx 10^5 cm s^-1`
C. `1 xx 10^11 cm s^-1`
D. `1.1 xx 10^9 cm s^-1`

Explanation

To solve the problem, we need to find the uncertainty in the electron's velocity (Δv) given the uncertainty in momentum (Δp) and the mass of the electron (m).

### Step-by-Step Solution:

1. **Identify the given values:**
- Uncertainty in momentum (Δp) = \(1 \times 10^{-18} \, \text{g cm s}^{-1}\)
- Mass of the electron (m) = \(9 \times 10^{-28} \, \text{g}\)

2. **Use the relationship between momentum and velocity:**
The momentum (p) of an object is given by the formula:
\[ p = m \cdot v \]
Therefore, the uncertainty in momentum (Δp) can be expressed as:
\[ \Delta p = m \cdot \Delta v \]
Rearranging this gives us:
\[ \Delta v = \frac{\Delta p}{m} \]

3. **Substitute the known values into the equation:**
\[ \Delta v = \frac{1 \times 10^{-18} \, \text{g cm s}^{-1}}{9 \times 10^{-28} \, \text{g}} \]

4. **Perform the calculation:**
\[ \Delta v = \frac{1 \times 10^{-18}}{9 \times 10^{-28}} = \frac{1}{9} \times 10^{10} \, \text{cm s}^{-1} \]
\[ \Delta v \approx 1.11 \times 10^{9} \, \text{cm s}^{-1} \]

5. **Final result:**
The uncertainty in the electron's velocity (Δv) is approximately:
\[ \Delta v \approx 1.11 \times 10^{9} \, \text{cm s}^{-1} \]