Difference between `n^(th)` and `(n+1)^(th)` Bohr's radius of H atom is equal to it's `(n-1)^(th)` Bohr's radius. The value of n is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the value of \( n \) such that the difference between the \( n^{th} \) and \( (n+1)^{th} \) Bohr radii of the hydrogen atom is equal to the \( (n-1)^{th} \) Bohr radius.

### Step-by-Step Solution:

1. **Write the formula for Bohr radius**:
The Bohr radius for the hydrogen atom is given by:
\[ r_n = 0.529 \frac{n^2}{z^2} \]
For hydrogen, \( z = 1 \), so:
\[ r_n = 0.529 n^2 \]

2. **Calculate the \( n^{th} \) and \( (n+1)^{th} \) Bohr radii**:
- The \( n^{th} \) Bohr radius:
\[ r_n = 0.529 n^2 \]
- The \( (n+1)^{th} \) Bohr radius:
\[ r_{n+1} = 0.529 (n+1)^2 = 0.529 (n^2 + 2n + 1) \]

3. **Find the difference between \( r_{n+1} \) and \( r_n \)**:
\[ r_{n+1} - r_n = 0.529 (n^2 + 2n + 1) - 0.529 n^2 \]
Simplifying this:
\[ r_{n+1} - r_n = 0.529 (2n + 1) \]

4. **Calculate the \( (n-1)^{th} \) Bohr radius**:
\[ r_{n-1} = 0.529 (n-1)^2 = 0.529 (n^2 - 2n + 1) \]

5. **Set up the equation**:
According to the problem, the difference between the \( n^{th} \) and \( (n+1)^{th} \) Bohr radii is equal to the \( (n-1)^{th} \) Bohr radius:
\[ 0.529 (2n + 1) = 0.529 (n^2 - 2n + 1) \]

6. **Cancel out the common factor**:
Since \( 0.529 \) is common on both sides, we can cancel it out:
\[ 2n + 1 = n^2 - 2n + 1 \]

7. **Rearrange the equation**:
Rearranging gives:
\[ n^2 - 4n = 0 \]

8. **Factor the equation**:
Factoring out \( n \):
\[ n(n - 4) = 0 \]

9. **Solve for \( n \)**:
The solutions are:
\[ n = 0 \quad \text{or} \quad n = 4 \]
Since \( n \) must be a positive integer (as it represents the principal quantum number), we discard \( n = 0 \).

10. **Final answer**:
Thus, the value of \( n \) is:
\[ n = 4 \]