The differnce between nth and `(n+1)` the Bohr radius of B atom is equal to be its `(n-1)` th Bohr 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 radius is equal to the \( (n-1) \)-th Bohr radius.

### Step-by-Step Solution:

1. **Understand the Bohr Radius Formula**:
The radius of the \( n \)-th orbit in a hydrogen-like atom is given by the formula:
\[ r_n = n^2 \cdot r_0 \]
where \( r_0 \) is the Bohr radius constant.

2. **Write the Expression for \( n \)-th and \( (n+1) \)-th Radius**:
- The radius of the \( n \)-th orbit is:
\[ r_n = n^2 \cdot r_0 \]
- The radius of the \( (n+1) \)-th orbit is:
\[ r_{n+1} = (n+1)^2 \cdot r_0 \]

3. **Calculate the Difference**:
The difference between the \( (n+1) \)-th and \( n \)-th radius is:
\[ r_{n+1} - r_n = (n+1)^2 \cdot r_0 - n^2 \cdot r_0 \]
Factoring out \( r_0 \):
\[ r_{n+1} - r_n = r_0 \left( (n+1)^2 - n^2 \right) \]

4. **Simplify the Difference**:
Expanding \( (n+1)^2 \):
\[ (n+1)^2 = n^2 + 2n + 1 \]
Therefore:
\[ r_{n+1} - r_n = r_0 \left( n^2 + 2n + 1 - n^2 \right) = r_0 (2n + 1) \]

5. **Write the Expression for \( (n-1) \)-th Radius**:
The radius of the \( (n-1) \)-th orbit is:
\[ r_{n-1} = (n-1)^2 \cdot r_0 \]

6. **Set Up the Equation**:
According to the problem, the difference between the \( n \)-th and \( (n+1) \)-th radius is equal to the \( (n-1) \)-th radius:
\[ r_0 (2n + 1) = (n-1)^2 \cdot r_0 \]

7. **Cancel \( r_0 \)** (assuming \( r_0 \neq 0 \)):
\[ 2n + 1 = (n-1)^2 \]

8. **Expand and Rearrange the Equation**:
Expanding \( (n-1)^2 \):
\[ 2n + 1 = n^2 - 2n + 1 \]
Rearranging gives:
\[ n^2 - 4n = 0 \]

9. **Factor the Quadratic**:
\[ n(n - 4) = 0 \]
This gives us two solutions:
\[ n = 0 \quad \text{or} \quad n = 4 \]

10. **Select the Valid Solution**:
Since \( n \) represents the principal quantum number, it must be a positive integer. Thus, we have:
\[ n = 4 \]

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