In an hydrogen atom, the electron revolves around the nucleus 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`

Explanation

To find the electrical potential produced by the nucleus (proton) at the position of the electron in a hydrogen atom, we can use the formula for electric potential \( V \) due to a point charge:

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

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 (which is the charge of the proton, \( +e \)),
- \( r \) is the distance from the nucleus to the electron (the radius of the orbit).

### Step-by-Step Solution:

1. **Identify the values:**
- The charge of the proton \( Q = +e = 1.6 \times 10^{-19} \, \text{C} \).
- The radius \( r = 0.53 \times 10^{-10} \, \text{m} \).
- Coulomb's constant \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

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:**
\[ 9 \times 10^9 \cdot 1.6 \times 10^{-19} = 14.4 \times 10^{-10} \, \text{N m}^2/\text{C} \]

4. **Calculate the potential:**
\[ V = \frac{14.4 \times 10^{-10}}{0.53 \times 10^{-10}} = \frac{14.4}{0.53} \, \text{V} \]

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

Thus, the electrical potential produced by the nucleus at the position of the electron is approximately **27.2 volts**.