In the Bohr model of the hydrogen atom an electron is pictured rotating in a circle (with a radius of `0.5xx10^(-10)m`) about the positive nucleus of the atom. The centripetal force is furnished by the electric attraction of the positive nucleus for the negative electron. How large is this force if the electron is moving with a speed of `2.3xx10^(6)`m/s?
(The mass of an electron is `9xx10^(-31)kg`)

Answer

To solve the problem, we need to calculate the centripetal force acting on the electron as it moves in a circular path around the nucleus of the hydrogen atom. The centripetal force can be calculated using the formula:

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

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 path.

### Step 1: Identify the given values
- Mass of the electron, \( m = 9 \times 10^{-31} \, \text{kg} \)
- Speed of the electron, \( v = 2.3 \times 10^{6} \, \text{m/s} \)
- Radius of the circular path, \( r = 0.5 \times 10^{-10} \, \text{m} \)

### Step 2: Substitute the values into the centripetal force formula
Now we will substitute the values of \( m \), \( v \), and \( r \) into the formula for centripetal force:

\[ F_c = \frac{(9 \times 10^{-31} \, \text{kg}) \times (2.3 \times 10^{6} \, \text{m/s})^2}{0.5 \times 10^{-10} \, \text{m}} \]

### Step 3: Calculate \( v^2 \)
First, calculate \( v^2 \):

\[ v^2 = (2.3 \times 10^{6})^2 = 5.29 \times 10^{12} \, \text{m}^2/\text{s}^2 \]

### Step 4: Substitute \( v^2 \) back into the formula
Now substitute \( v^2 \) back into the centripetal force equation:

\[ F_c = \frac{(9 \times 10^{-31}) \times (5.29 \times 10^{12})}{0.5 \times 10^{-10}} \]

### Step 5: Perform the multiplication in the numerator
Calculate the numerator:

\[ 9 \times 10^{-31} \times 5.29 \times 10^{12} = 4.761 \times 10^{-18} \, \text{kg m}^2/\text{s}^2 \]

### Step 6: Divide by the radius
Now divide by the radius:

\[ F_c = \frac{4.761 \times 10^{-18}}{0.5 \times 10^{-10}} = \frac{4.761 \times 10^{-18}}{5 \times 10^{-11}} = 9.522 \times 10^{-8} \, \text{N} \]

### Step 7: Final result
Thus, the magnitude of the centripetal force acting on the electron is:

\[ F_c \approx 9.5 \times 10^{-8} \, \text{N} \]