In a region of uniform electric field, as an electron travels from A to B, it slows from
`v_A = 6.1 xx 10^(6) m//s` to `v_(B) = 4.5 xx 10^(6) m//s`. The potential change
`Delta V = V_(B) - V_(A)` in volts nearly.

Options

A. 18
B. `-18`
C. `+48`
D. `-48`

Answer

To find the potential change \( \Delta V = V_B - V_A \) as the electron travels from point A to point B in a uniform electric field, we can use the principle of conservation of energy. The kinetic energy lost by the electron will be equal to the potential energy gained.

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial velocity \( v_A = 6.1 \times 10^6 \, \text{m/s} \)
- Final velocity \( v_B = 4.5 \times 10^6 \, \text{m/s} \)
- Mass of the electron \( m = 9.1 \times 10^{-31} \, \text{kg} \)

2. **Calculate the initial kinetic energy (KE) at point A:**
\[ KE_A = \frac{1}{2} m v_A^2 = \frac{1}{2} (9.1 \times 10^{-31}) (6.1 \times 10^6)^2 \]

3. **Calculate the final kinetic energy (KE) at point B:**
\[ KE_B = \frac{1}{2} m v_B^2 = \frac{1}{2} (9.1 \times 10^{-31}) (4.5 \times 10^6)^2 \]

4. **Calculate the change in kinetic energy (\( \Delta KE \)):**
\[ \Delta KE = KE_A - KE_B \]

5. **Substituting the values:**
\[ KE_A = \frac{1}{2} (9.1 \times 10^{-31}) (6.1^2 \times 10^{12}) = \frac{1}{2} (9.1 \times 10^{-31}) (37.21 \times 10^{12}) = 1.69 \times 10^{-18} \, \text{J} \]
\[ KE_B = \frac{1}{2} (9.1 \times 10^{-31}) (4.5^2 \times 10^{12}) = \frac{1}{2} (9.1 \times 10^{-31}) (20.25 \times 10^{12}) = 9.19 \times 10^{-19} \, \text{J} \]
\[ \Delta KE = 1.69 \times 10^{-18} - 9.19 \times 10^{-19} = 7.71 \times 10^{-19} \, \text{J} \]

6. **Relate the change in kinetic energy to the change in potential energy (\( \Delta U \)):**
\[ \Delta U = -\Delta KE \]

7. **Calculate the change in potential (\( \Delta V \)) using the relation \( \Delta U = q \Delta V \):**
- The charge of the electron \( q = -1.6 \times 10^{-19} \, \text{C} \)
\[ \Delta V = \frac{\Delta U}{q} = \frac{-\Delta KE}{-1.6 \times 10^{-19}} = \frac{7.71 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 48.19 \, \text{V} \]

8. **Determine the final result for \( \Delta V \):**
\[ \Delta V = V_B - V_A \approx -48.19 \, \text{V} \]

### Final Answer:
The potential change \( \Delta V = V_B - V_A \) is approximately \(-48.19 \, \text{V}\).