Question

If the difference between \((n+1)^{th}\) Bohr radius and \(n^{th}\) Bohr radius is given by \((n-1)^{th}\) Bohr radius, then the value of n is

• A. \(4\)
• B. \(3\)
• C. \(2\)
• D. \(1\)

Hint:

The Bohr radius of orbit n is proportional to n squared. Set (n+1)^2 - n^2 = (n-1)^2.

Answer 1

The Correct Option is A

Step 1: Plug in the option values:
Test each option with $r\propto n^2$.
$n = 4$: $5^2 - 4^2 = 9$ and $3^2 = 9$. It matches.
$n = 3$: $16 - 9 = 7$ but $2^2 = 4$. No.
$n = 2$: $9 - 4 = 5$ but $1^2 = 1$. No.
$n = 1$: $4 - 1 = 3$ but $0^2 = 0$. No.

Step 2: Algebra confirms:
$2n+1 = (n-1)^2$ simplifies to $n(n-4) = 0$.

Step 3: Why the other values fail:
For $n = 3$ the left side is $16 - 9 = 7$ while the right side is $4$. For $n = 2$ the left side is $5$ while the right side is $1$. For $n = 1$ the right side would involve the zeroth orbit, which does not exist. Only $n = 4$ gives equal sides, 9 and 9.

Final Answer:
Option (A). \[ \boxed{4 \text{ (A)}} \]