Given that `r_(n-1)+r_(n)=r_(n+1)` where r represents Bohr's orbit for `(n-1)^("th"),n^("th")` and `(n+1)^("th")` orbit, then n is

Answer

To solve the problem, we need to analyze the given equation involving the radii of the Bohr orbits:

**Given:**
\[ r_{n-1} + r_n = r_{n+1} \]

**Step 1: Write the formula for the Bohr radius.**
The radius of the \( n \)-th Bohr orbit is given by:
\[ r_n = \frac{0.529 \, n^2}{Z} \]
where \( Z \) is the atomic number. For simplicity, we will assume \( Z = 1 \) (for hydrogen).

**Step 2: Substitute the values of \( r_{n-1} \), \( r_n \), and \( r_{n+1} \) using the formula.**
Using the formula for the radii:
- \( r_{n-1} = \frac{0.529 \, (n-1)^2}{1} = 0.529 \, (n-1)^2 \)
- \( r_n = \frac{0.529 \, n^2}{1} = 0.529 \, n^2 \)
- \( r_{n+1} = \frac{0.529 \, (n+1)^2}{1} = 0.529 \, (n+1)^2 \)

**Step 3: Substitute these expressions into the given equation.**
Now substituting into the equation:
\[ 0.529 \, (n-1)^2 + 0.529 \, n^2 = 0.529 \, (n+1)^2 \]

**Step 4: Factor out the common term.**
We can factor out \( 0.529 \) from all terms:
\[ (n-1)^2 + n^2 = (n+1)^2 \]

**Step 5: Expand the squares.**
Now, we expand the squares:
\[ (n^2 - 2n + 1) + n^2 = (n^2 + 2n + 1) \]
This simplifies to:
\[ 2n^2 - 2n + 1 = n^2 + 2n + 1 \]

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

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

**Step 8: Solve for \( n \).**
Setting each factor to zero gives:
1. \( n = 0 \)
2. \( n - 4 = 0 \) → \( n = 4 \)

Since \( n \) represents the principal quantum number of an orbit, it cannot be zero. Therefore, the only acceptable solution is:
\[ n = 4 \]

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

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