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=2
B. n=3 to n=8
C. n=4
D. n=16

Answer

To find the principal quantum number \( n \) of the final state of the hydrogen atom after it has collided with an electron, we can use the relationship between the radius of the atom and the principal quantum number as given by the Bohr model.

### Step-by-step Solution:

1. **Understand the relationship between radius and principal quantum number**:
The radius \( r \) of a hydrogen atom in the nth state is given by the formula:
\[ r = 0.529 \times n^2 \text{ (in meters)} \]
Here, \( z \) (the atomic number) is 1 for hydrogen.

2. **Set up the ratio of initial and final radii**:
According to the Bohr model, the radius is directly proportional to the square of the principal quantum number:
\[ \frac{r_i}{r_f} = \frac{n_i^2}{n_f^2} \]
Where:
- \( r_i = 5.3 \times 10^{-11} \, m \) (initial radius)
- \( r_f = 21.2 \times 10^{-11} \, m \) (final radius)
- \( n_i = 1 \) (initial state, ground state)

3. **Substitute the values into the equation**:
Substitute the known values into the equation:
\[ \frac{5.3 \times 10^{-11}}{21.2 \times 10^{-11}} = \frac{1^2}{n_f^2} \]

4. **Simplify the equation**:
The \( 10^{-11} \) terms cancel out:
\[ \frac{5.3}{21.2} = \frac{1}{n_f^2} \]

5. **Calculate the left side**:
Calculate \( \frac{5.3}{21.2} \):
\[ \frac{5.3}{21.2} = 0.25 \]
Therefore, we have:
\[ 0.25 = \frac{1}{n_f^2} \]

6. **Invert the equation**:
Inverting both sides gives:
\[ n_f^2 = \frac{1}{0.25} = 4 \]

7. **Solve for \( n_f \)**:
Taking the square root of both sides:
\[ n_f = 2 \]

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