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. `1xx10^(5)"cm s"^(-1)`
B. `1xx10^(11)"cm s"^(-1)`
C. `1xx10^(9)"cm s"^(-1)`
D. `1xx10^(6)"cm s"^(-1)`

Explanation

To solve the problem, we need to find the uncertainty in the electron's velocity given the uncertainty in its momentum and its mass.

### Step-by-Step Solution:

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

2. **Understand the relationship between momentum, mass, and velocity:**
- The momentum (P) of an object is given by the formula:
\[ P = m \cdot v \]
- Where \(v\) is the velocity.

3. **Express the uncertainty in momentum in terms of uncertainty in velocity:**
- The uncertainty in momentum can also be expressed as:
\[ \Delta P = m \cdot \Delta v \]
- Here, \(\Delta v\) is the uncertainty in velocity.

4. **Rearranging the equation to find the uncertainty in velocity:**
- We can rearrange the equation to solve for \(\Delta v\):
\[ \Delta v = \frac{\Delta P}{m} \]

5. **Substituting the known values:**
- Now, substituting the values of \(\Delta P\) and \(m\):
\[ \Delta v = \frac{1 \times 10^{-18} \, \text{g cm/s}}{9 \times 10^{-28} \, \text{g}} \]

6. **Calculating the uncertainty in velocity:**
- Performing the division:
\[ \Delta v = \frac{1}{9} \times 10^{10} \, \text{cm/s} \]
- This simplifies to:
\[ \Delta v \approx 0.1111 \times 10^{10} \, \text{cm/s} \]
- Which can be expressed as:
\[ \Delta v \approx 1.11 \times 10^{9} \, \text{cm/s} \]

7. **Final Result:**
- Therefore, the uncertainty in the electron's velocity is approximately:
\[ \Delta v \approx 1.1 \times 10^{9} \, \text{cm/s} \]