To determine the speed at which an electron should revolve around the nucleus of a hydrogen atom so that it is not pulled into the nucleus by electrostatic attraction, we can follow these steps:
### Step 1: Understand the Forces Involved
The electron in orbit around the nucleus experiences two forces:
- **Electrostatic Force (Fe)**: This is the force of attraction between the positively charged nucleus and the negatively charged electron.
- **Centripetal Force (Fc)**: This is the force required to keep the electron moving in a circular path.
For the electron to remain in stable orbit, these two forces must be equal:
\[ F_e = F_c \]
### Step 2: Write the Expressions for the Forces
1. **Electrostatic Force** is given by Coulomb's Law:
\[
F_e = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Ze^2}{r^2}
\]
where:
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( e \) is the charge of the electron (\( e = 1.6 \times 10^{-19} \, C \)),
- \( r \) is the radius of the orbit.
2. **Centripetal Force** is given by:
\[
F_c = \frac{mv^2}{r}
\]
where:
- \( m \) is the mass of the electron (\( m = 9.1 \times 10^{-31} \, kg \)),
- \( v \) is the speed of the electron.
### Step 3: Set the Forces Equal
Setting the two forces equal gives us:
\[
\frac{1}{4 \pi \epsilon_0} \cdot \frac{Ze^2}{r^2} = \frac{mv^2}{r}
\]
### Step 4: Simplify the Equation
Since \( Z = 1 \) for hydrogen, we can simplify:
\[
\frac{1}{4 \pi \epsilon_0} \cdot \frac{e^2}{r^2} = \frac{mv^2}{r}
\]
Multiplying both sides by \( r \) gives:
\[
\frac{1}{4 \pi \epsilon_0} \cdot \frac{e^2}{r} = mv^2
\]
### Step 5: Solve for \( v^2 \)
Rearranging the equation to solve for \( v^2 \):
\[
v^2 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{e^2}{mr}
\]
### Step 6: Substitute Known Values
Substituting the known values:
- \( e = 1.6 \times 10^{-19} \, C \)
- \( \epsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \)
- \( m = 9.1 \times 10^{-31} \, kg \)
- \( r = 0.5 \, Å = 0.5 \times 10^{-10} \, m = 5 \times 10^{-11} \, m \)
Calculating \( v^2 \):
\[
v^2 = \frac{(1.6 \times 10^{-19})^2}{4 \pi (8.85 \times 10^{-12}) (9.1 \times 10^{-31}) (5 \times 10^{-11})}
\]
### Step 7: Calculate the Numerical Value
Calculating the above expression step by step:
1. Calculate \( (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \).
2. Calculate \( 4 \pi (8.85 \times 10^{-12}) = 1.112 \times 10^{-10} \).
3. Calculate \( 9.1 \times 10^{-31} \times 5 \times 10^{-11} = 4.55 \times 10^{-41} \).
4. Now, substituting these values into the equation for \( v^2 \):
\[
v^2 = \frac{2.56 \times 10^{-38}}{1.112 \times 10^{-10} \times 4.55 \times 10^{-41}} \approx 2.25 \times 10^{6} \, m^2/s^2
\]
### Step 8: Take the Square Root to Find \( v \)
Taking the square root gives:
\[
v \approx \sqrt{2.25 \times 10^{6}} \approx 1500 \, m/s
\]
### Final Answer
The speed at which the electron should revolve around the nucleus of a hydrogen atom is approximately:
\[
v \approx 2.25 \times 10^{6} \, m/s
\]
