The speed of the electron in the hydrogen atom is approximately `2.2 xx 10^(6) m//s`. What is the centripetal force acting on the electron ifthe radius of the circular orbit is `0.53 xx 10^(-10) m`?Mass of the electron is `9.1 xx 10^(-31) kg`

Explanation

To find the centripetal force acting on the electron in a hydrogen atom, we will use the formula for centripetal force:

\[ F_c = \frac{mv^2}{r} \]

where:
- \( F_c \) is the centripetal force,
- \( m \) is the mass of the electron,
- \( v \) is the speed of the electron,
- \( r \) is the radius of the circular orbit.

### Step-by-Step Solution:

1. **Identify the given values:**
- Speed of the electron, \( v = 2.2 \times 10^6 \, \text{m/s} \)
- Radius of the circular orbit, \( r = 0.53 \times 10^{-10} \, \text{m} \)
- Mass of the electron, \( m = 9.1 \times 10^{-31} \, \text{kg} \)

2. **Substitute the values into the centripetal force formula:**
\[ F_c = \frac{(9.1 \times 10^{-31} \, \text{kg}) \cdot (2.2 \times 10^6 \, \text{m/s})^2}{0.53 \times 10^{-10} \, \text{m}} \]

3. **Calculate \( v^2 \):**
\[ v^2 = (2.2 \times 10^6)^2 = 4.84 \times 10^{12} \, \text{m}^2/\text{s}^2 \]

4. **Calculate the numerator:**
\[ \text{Numerator} = 9.1 \times 10^{-31} \, \text{kg} \cdot 4.84 \times 10^{12} \, \text{m}^2/\text{s}^2 = 4.40 \times 10^{-18} \, \text{kg m}^2/\text{s}^2 \]

5. **Calculate the centripetal force:**
\[ F_c = \frac{4.40 \times 10^{-18} \, \text{kg m}^2/\text{s}^2}{0.53 \times 10^{-10} \, \text{m}} = 8.32 \times 10^{-8} \, \text{N} \]

### Final Answer:
The centripetal force acting on the electron is approximately \( 8.32 \times 10^{-8} \, \text{N} \).