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

Explanation

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's radius of a hydrogen atom is equal to its \( (n-1)^{th} \) Bohr's radius.

### Step-by-step Solution:

1. **Bohr's Radius Formula**:
The formula for the radius of the \( n^{th} \) orbit in a 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. **Difference Between Radii**:
We need to find the difference between the \( n^{th} \) and \( (n+1)^{th} \) radii:
\[ r_{n+1} - r_n = 0.529 (n+1)^2 - 0.529 n^2 \]
Simplifying this:
\[ r_{n+1} - r_n = 0.529 \left( (n^2 + 2n + 1) - n^2 \right) = 0.529 (2n + 1) \]

3. **Setting Up the Equation**:
According to the problem, this difference is equal to the \( (n-1)^{th} \) radius:
\[ r_{n+1} - r_n = r_{n-1} \]
Therefore:
\[ 0.529 (2n + 1) = 0.529 (n-1)^2 \]

4. **Cancelling Out the Common Factor**:
We can cancel \( 0.529 \) from both sides:
\[ 2n + 1 = (n-1)^2 \]

5. **Expanding the Right Side**:
Expanding \( (n-1)^2 \):
\[ 2n + 1 = n^2 - 2n + 1 \]

6. **Rearranging the Equation**:
Rearranging gives us:
\[ n^2 - 2n + 1 - 2n - 1 = 0 \]
Simplifying further:
\[ n^2 - 4n = 0 \]

7. **Factoring the Equation**:
Factoring out \( n \):
\[ n(n - 4) = 0 \]

8. **Finding the Values of n**:
This gives us two possible solutions:
\[ n = 0 \quad \text{or} \quad n = 4 \]

9. **Determining the Valid Solution**:
Since \( n \) represents the principal quantum number (which must be a positive integer), we discard \( n = 0 \). Thus, the only valid solution is:
\[ n = 4 \]

### Final Answer:
The value of \( n \) is \( 4 \).