The radius of hydrogen atom in its ground state is `5.3 xx 10^(-11)m`. After collision with an electron it is found to have a radius of `21.2 xx 10^(-11)m`. What is the principle quantum number of `n` of the final state of the atom ?

Options

A. n = 4
B. n = 2
C. n = 16
D. n = 3

Answer

To find the principal quantum number \( n \) of the final state of the hydrogen atom after a collision with an electron, we can use the formula that relates the radius of the hydrogen atom in different energy levels:

\[ R_n = R_1 \cdot \frac{n^2}{Z} \]

Where:
- \( R_n \) is the radius of the atom in the \( n \)-th state.
- \( R_1 \) is the radius of the atom in the ground state (first state).
- \( n \) is the principal quantum number.
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)).

### Given:
- \( R_1 = 5.3 \times 10^{-11} \, m \) (radius in the ground state)
- \( R_n = 21.2 \times 10^{-11} \, m \) (radius after collision)

### Steps to Solve:

1. **Write down the formula:**
\[ R_n = R_1 \cdot \frac{n^2}{Z} \]

2. **Substitute the known values into the formula:**
Since \( Z = 1 \) for hydrogen, we have:
\[ 21.2 \times 10^{-11} = 5.3 \times 10^{-11} \cdot n^2 \]

3. **Rearranging the equation to solve for \( n^2 \):**
\[ n^2 = \frac{21.2 \times 10^{-11}}{5.3 \times 10^{-11}} \]

4. **Calculate the right side:**
\[ n^2 = \frac{21.2}{5.3} \]
Performing the division:
\[ n^2 = 4 \]

5. **Taking the square root to find \( n \):**
\[ n = \sqrt{4} = 2 \]

### Conclusion:
The principal quantum number \( n \) of the final state of the hydrogen atom is \( n = 2 \).