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. 4
B. 3
C. 2
D. 1

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

### Step-by-step Solution:

1. **Write the formula for Bohr's radius**:
The Bohr's 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's radii**:
- For \( n^{th} \) radius:
\[ R_n = 0.529 n^2 \]
- For \( (n+1)^{th} \) 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. **Set the difference equal to \( R_{n-1} \)**:
- For \( (n-1)^{th} \) radius:
\[ R_{n-1} = 0.529 (n-1)^2 = 0.529 (n^2 - 2n + 1) \]
- Now, we set up the equation:
\[ 0.529 (2n + 1) = 0.529 (n^2 - 2n + 1) \]

5. **Cancel \( 0.529 \) from both sides**:
\[ 2n + 1 = n^2 - 2n + 1 \]

6. **Rearrange the equation**:
\[ n^2 - 2n - 2n + 1 - 1 = 0 \]
This simplifies to:
\[ n^2 - 4n = 0 \]

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

8. **Find the possible values of \( n \)**:
From the factored equation, we get:
\[ n = 0 \quad \text{or} \quad n = 4 \]
Since \( n \) cannot be zero (as it represents the principal quantum number), we conclude:
\[ n = 4 \]

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