Q148 / 150 · MHT CET 2026 (18 April Shift 2) · Physics
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
- Correct answerA$4$
- B$3$
- C$2$
- D$1$
Solution
The radius of the $n^{th}$ Bohr orbit is given by $r_n = r_0 n^2$, where $r_0$ is the radius of the first Bohr orbit.
According to the given condition:
$r_{n+1} - r_n = r_{n-1}$
Substituting $r_n = r_0 n^2$:
$r_0 (n+1)^2 - r_0 n^2 = r_0 (n-1)^2$
Dividing both sides by $r_0$:
$(n+1)^2 - n^2 = (n-1)^2$
$n^2 + 2n + 1 - n^2 = n^2 - 2n + 1$
$2n + 1 = n^2 - 2n + 1$
$n^2 - 4n = 0$
$n(n - 4) = 0$
Since $n$ must be a positive integer, $n = 4$.
According to the given condition:
$r_{n+1} - r_n = r_{n-1}$
Substituting $r_n = r_0 n^2$:
$r_0 (n+1)^2 - r_0 n^2 = r_0 (n-1)^2$
Dividing both sides by $r_0$:
$(n+1)^2 - n^2 = (n-1)^2$
$n^2 + 2n + 1 - n^2 = n^2 - 2n + 1$
$2n + 1 = n^2 - 2n + 1$
$n^2 - 4n = 0$
$n(n - 4) = 0$
Since $n$ must be a positive integer, $n = 4$.
Hence, the correct answer is option: 'A'.