In an hydrogen atom, the electron revolves around the nucles in an orbit of radius `0.53 xx 10^(-10) m`. Then the electrical potential produced by the nucleus at the position of the electron is

Options

A. `-13.6 V`
B. `-27.2 V`
C. `27.2 V`
D. `13.6 V`

Answer

To find the electrical potential produced by the nucleus at the position of the electron in a hydrogen atom, we will use the formula for electric potential \( V \):

\[ V = \frac{k \cdot q}{r} \]

where:
- \( V \) is the electric potential,
- \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \),
- \( q \) is the charge of the nucleus (for hydrogen, this is the charge of a proton, approximately \( 1.6 \times 10^{-19} \, \text{C} \)),
- \( r \) is the radius of the orbit of the electron, given as \( 0.53 \times 10^{-10} \, \text{m} \).

### Step-by-step Solution:

1. **Identify the values**:
- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( q = 1.6 \times 10^{-19} \, \text{C} \)
- \( r = 0.53 \times 10^{-10} \, \text{m} \)

2. **Substitute the values into the formula**:
\[ V = \frac{(9 \times 10^9) \cdot (1.6 \times 10^{-19})}{0.53 \times 10^{-10}} \]

3. **Calculate the numerator**:
\[ 9 \times 10^9 \cdot 1.6 \times 10^{-19} = 14.4 \times 10^{-10} \]

4. **Now substitute this back into the equation**:
\[ V = \frac{14.4 \times 10^{-10}}{0.53 \times 10^{-10}} \]

5. **Simplify the equation**:
\[ V = \frac{14.4}{0.53} \]

6. **Perform the division**:
\[ V \approx 27.2 \, \text{V} \]

### Final Answer:
The electrical potential produced by the nucleus at the position of the electron is approximately **27.2 volts**.