At what speed should the electron revolve around the nucleus of a hydrogen atom in order that it may not be pulled into the nucleus by electrostatic attraction ? Take the radius of orbit of an electron as `0.5Å` , the mass of the electron as `9.1xx10^(-31)` kg and charge as `1.6xx10^(-19)C`.

Options

A. `2.25xx10^(4) ms ^(-1)`
B. `2.25xx10^(5) ms ^(-1)`
C. `2.25xx10^(6) ms ^(-1)`
D. `2.25xx10^(7) ms ^(-1)`

Answer

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 is in a circular orbit around the nucleus due to the balance between the electrostatic force of attraction and the centripetal force required to keep the electron in circular motion.

### Step 2: Write the Formula for Electrostatic Force
The electrostatic force \( F \) between the electron and the proton can be expressed using Coulomb's law:
\[ F = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{r^2} \]
where:
- \( q_1 \) is the charge of the proton \( (1.6 \times 10^{-19} \, C) \)
- \( q_2 \) is the charge of the electron \( (-1.6 \times 10^{-19} \, C) \)
- \( r \) is the radius of the orbit \( (0.5 \, \text{Å} = 0.5 \times 10^{-10} \, m) \)
- \( \epsilon_0 \) is the permittivity of free space \( (8.85 \times 10^{-12} \, C^2/(N \cdot m^2)) \)

### Step 3: Write the Formula for Centripetal Force
The centripetal force \( F_c \) required to keep the electron in circular motion is given by:
\[ F_c = \frac{m v^2}{r} \]
where:
- \( m \) is the mass of the electron \( (9.1 \times 10^{-31} \, kg) \)
- \( v \) is the speed of the electron
- \( r \) is the radius of the orbit

### Step 4: Set the Forces Equal
For the electron to remain in orbit without falling into the nucleus, the electrostatic force must equal the centripetal force:
\[ \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{r^2} = \frac{m v^2}{r} \]

### Step 5: Rearrange the Equation to Solve for \( v \)
Rearranging the equation gives:
\[ v^2 = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{m} \cdot \frac{1}{r} \]
Taking the square root:
\[ v = \sqrt{\frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{m} \cdot \frac{1}{r}} \]

### Step 6: Substitute the Known Values
Substituting the known values into the equation:
- \( q_1 = 1.6 \times 10^{-19} \, C \)
- \( q_2 = -1.6 \times 10^{-19} \, C \)
- \( r = 0.5 \times 10^{-10} \, m \)
- \( m = 9.1 \times 10^{-31} \, kg \)
- \( \epsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \)

Calculating the values:
\[ v = \sqrt{\frac{1}{4\pi (8.85 \times 10^{-12})} \cdot \frac{(1.6 \times 10^{-19})^2}{(9.1 \times 10^{-31})} \cdot \frac{1}{(0.5 \times 10^{-10})}} \]

### Step 7: Calculate the Speed
After performing the calculations, we find:
\[ v \approx 2.25 \times 10^6 \, m/s \]

### Conclusion
The speed at which the electron should revolve around the nucleus of a hydrogen atom to avoid being pulled into the nucleus is approximately \( 2.25 \times 10^6 \, m/s \).

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