An electron is liberated from a hot filament and attracted by an anode at 1200 V. What is its speed when it strike the anode?(e=`1.6xx10^(-19)`C)

Answer

To find the speed of the electron when it strikes the anode at 1200 V, we can use the principle of conservation of energy. The gain in kinetic energy of the electron will be equal to the loss in potential energy as it moves from the filament to the anode.

### Step-by-step Solution:

1. **Identify the initial and final energies**:
- The electron is initially at rest when it is liberated from the hot filament, so its initial kinetic energy (KE_initial) is 0.
- The potential energy (PE_initial) at the filament is negative because it is in an electric field, but we can consider it as a reference point.

2. **Calculate the potential energy at the anode**:
- The potential energy at the anode (PE_final) can be calculated using the formula:
\[ PE = V \cdot q \]
where \( V = 1200 \, \text{V} \) (the voltage of the anode) and \( q = -1.6 \times 10^{-19} \, \text{C} \) (the charge of the electron).
- Thus, the potential energy at the anode is:
\[ PE_{final} = 1200 \, \text{V} \cdot (-1.6 \times 10^{-19} \, \text{C}) = -1.92 \times 10^{-16} \, \text{J} \]

3. **Calculate the change in potential energy**:
- The change in potential energy (ΔPE) from the filament to the anode is:
\[ \Delta PE = PE_{final} - PE_{initial} \]
- Since we can consider the potential energy at the filament to be 0 (as a reference), we have:
\[ \Delta PE = -1.92 \times 10^{-16} \, \text{J} - 0 = -1.92 \times 10^{-16} \, \text{J} \]

4. **Relate the change in potential energy to kinetic energy**:
- The gain in kinetic energy (KE) is equal to the loss in potential energy:
\[ KE = -\Delta PE = 1.92 \times 10^{-16} \, \text{J} \]

5. **Use the kinetic energy formula to find speed**:
- The kinetic energy of the electron can also be expressed as:
\[ KE = \frac{1}{2} m v^2 \]
where \( m = 9.11 \times 10^{-31} \, \text{kg} \) (mass of the electron) and \( v \) is the speed we want to find.
- Setting the two expressions for kinetic energy equal gives:
\[ 1.92 \times 10^{-16} = \frac{1}{2} (9.11 \times 10^{-31}) v^2 \]

6. **Solve for \( v^2 \)**:
- Rearranging the equation gives:
\[ v^2 = \frac{2 \cdot 1.92 \times 10^{-16}}{9.11 \times 10^{-31}} \]

7. **Calculate \( v \)**:
- Performing the calculation:
\[ v^2 = \frac{3.84 \times 10^{-16}}{9.11 \times 10^{-31}} \approx 4.22 \times 10^{14} \]
- Taking the square root:
\[ v \approx \sqrt{4.22 \times 10^{14}} \approx 2.06 \times 10^7 \, \text{m/s} \]

### Final Answer:
The speed of the electron when it strikes the anode is approximately \( 2.06 \times 10^7 \, \text{m/s} \).