The difference between the n^{th} and (n+1)^{ | vedclass

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(D) The Bohr radius for the $n^{th}$ orbit is given by $r_n = a_0 n^2$,where $a_0$ is the Bohr radius constant.According to the problem,the difference

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$4$

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(D) The Bohr radius for the $n^{th}$ orbit is given by $r_n = a_0 n^2$,where $a_0$ is the Bohr radius constant.According to the problem,the difference between the $(n+1)^{th}$ and $n^{th}$ radius is equal to the $(n-1)^{th}$ radius:$r_{n+1} - r_n = r_{n-1}$Substituting the formula $r_n = a_0 n^2$:$a_0(n+1)^2 - a_0 n^2 = a_0(n-1)^2$Dividing by $a_0$:$(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 greater than $1$ for the $(n-1)^{th}$ orbit to exist,we have $n = 4$.

The difference between the $n^{th}$ and $(n+1)^{th}$ Bohr radius of an $H$ atom is equal to its $(n-1)^{th}$ Bohr radius. The value of $n$ is:

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