The mass of the electrons `9.8 xx 10^(-28)` gram and uncertainty in the velocity equal to `2 xx 10^(-3) cm//sec`. The uncertainty in the position of an electron is `(h = 6.62 xx 10^(-27)` ergsec)

Options

A. `2.9 xx 10^(+2)cm`
B. `2.9 xx 10^(-2)cm`
C. `2.9 xx 10^(-12)cm^(-1)`
D. `2.9 xx 10^(+12)cm^(-1)`

Explanation

To find the uncertainty in the position of an electron given its mass and uncertainty in velocity, we can use the Heisenberg Uncertainty Principle. The principle states that:

\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]

Where:
- \(\Delta x\) is the uncertainty in position,
- \(\Delta p\) is the uncertainty in momentum,
- \(h\) is Planck's constant.

1. **Identify the given values:**
- Mass of the electron, \(m = 9.8 \times 10^{-28}\) g
- Uncertainty in velocity, \(\Delta v = 2 \times 10^{-3}\) cm/s
- Planck's constant, \(h = 6.62 \times 10^{-27}\) erg·s

2. **Convert Planck's constant to Joules:**
- We know that \(1 \text{ erg} = 10^{-7} \text{ Joules}\).
- Therefore, \(h = 6.62 \times 10^{-27} \text{ erg·s} = 6.62 \times 10^{-27} \times 10^{-7} \text{ Joules·s} = 6.62 \times 10^{-34} \text{ Joules·s}\).

3. **Calculate the uncertainty in momentum (\(\Delta p\)):**
- The uncertainty in momentum is given by:
\[ \Delta p = m \cdot \Delta v \]
- Substituting the values:
\[ \Delta p = (9.8 \times 10^{-28} \text{ g}) \cdot (2 \times 10^{-3} \text{ cm/s}) \]
- Convert mass from grams to kilograms:
\[ 9.8 \times 10^{-28} \text{ g} = 9.8 \times 10^{-31} \text{ kg} \]
- Now calculate \(\Delta p\):
\[ \Delta p = (9.8 \times 10^{-31} \text{ kg}) \cdot (2 \times 10^{-5} \text{ m/s}) = 1.96 \times 10^{-35} \text{ kg·m/s} \]

4. **Substitute into the uncertainty principle equation:**
\[ \Delta x \cdot \Delta p = \frac{h}{4\pi} \]
- Rearranging gives:
\[ \Delta x = \frac{h}{4\pi \Delta p} \]

5. **Calculate \(\Delta x\):**
- Substitute the values:
\[ \Delta x = \frac{6.62 \times 10^{-34} \text{ J·s}}{4 \cdot 3.14 \cdot 1.96 \times 10^{-35} \text{ kg·m/s}} \]
- Calculate the denominator:
\[ 4 \cdot 3.14 \cdot 1.96 \approx 24.66 \]
- Now calculate \(\Delta x\):
\[ \Delta x = \frac{6.62 \times 10^{-34}}{24.66 \times 10^{-35}} \approx 2.68 \text{ cm} \]

6. **Final result:**
- The uncertainty in the position of the electron is approximately \(2.68 \text{ cm}\).