At what speed must an electron revolve around the nucleus of hydrogen atom so that it may not be pulled into the nucleus by electrostatic attraction? Given, mass of electron `=9.1xx10^(-31) kg`, radius of orbit `=0.5xx10^(-10)m and e=1.6xx10^(-19)C.`

Answer

To find the speed at which an electron must revolve around the nucleus of a hydrogen atom so that it is not pulled into the nucleus by electrostatic attraction, we can equate the electrostatic force to the centripetal force acting on the electron.

### Step-by-Step Solution:

1. **Identify the Forces**:
- The electrostatic force \( F_e \) between the electron and the nucleus (proton) is given by Coulomb's Law:
\[ F_e = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]
where \( k \) is Coulomb's constant (\( k \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)), \( q_1 \) and \( q_2 \) are the charges of the electron and proton (both equal to \( e = 1.6 \times 10^{-19} \, \text{C} \)), and \( r \) is the radius of the orbit.

2. **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 (\( m = 9.1 \times 10^{-31} \, \text{kg} \)) and \( v \) is the speed of the electron.

3. **Set the Forces Equal**:
- For the electron to remain in orbit without spiraling into the nucleus, the electrostatic force must equal the centripetal force:
\[ F_e = F_c \]
Therefore:
\[ \frac{k \cdot e^2}{r^2} = \frac{m v^2}{r} \]

4. **Rearranging the Equation**:
- Multiply both sides by \( r^2 \) to eliminate the denominator:
\[ k \cdot e^2 = m v^2 \cdot r \]
- Now, solve for \( v^2 \):
\[ v^2 = \frac{k \cdot e^2}{m \cdot r} \]

5. **Substituting Values**:
- Substitute the known values into the equation:
- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)
- \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- \( r = 0.5 \times 10^{-10} \, \text{m} \)

\[ v^2 = \frac{(9 \times 10^9) \cdot (1.6 \times 10^{-19})^2}{(9.1 \times 10^{-31}) \cdot (0.5 \times 10^{-10})} \]

6. **Calculating \( v^2 \)**:
- Calculate \( e^2 \):
\[ e^2 = (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 \]
- Now substitute:
\[ v^2 = \frac{(9 \times 10^9) \cdot (2.56 \times 10^{-38})}{(9.1 \times 10^{-31}) \cdot (0.5 \times 10^{-10})} \]

7. **Final Calculation**:
- Calculate the numerator:
\[ 9 \times 10^9 \cdot 2.56 \times 10^{-38} = 2.304 \times 10^{-28} \]
- Calculate the denominator:
\[ 9.1 \times 10^{-31} \cdot 0.5 \times 10^{-10} = 4.55 \times 10^{-41} \]
- Now calculate \( v^2 \):
\[ v^2 = \frac{2.304 \times 10^{-28}}{4.55 \times 10^{-41}} \approx 5.06 \times 10^{12} \]
- Finally, take the square root to find \( v \):
\[ v \approx \sqrt{5.06 \times 10^{12}} \approx 2.25 \times 10^6 \, \text{m/s} \]

### Conclusion:
The speed at which an electron must revolve around the nucleus of a hydrogen atom is approximately \( 2.25 \times 10^6 \, \text{m/s} \).