I am trying to verify the following claim in my textbook,

The frequency of radiation of emitted when a hydrogen atom de-excites from level $n$ to $n-1$. For large $n$ this frequency equals the classical frequency of revolution of electron in the orbit.

Classical frequency is $$\nu = \dfrac{v}{2\pi r} = 2Rc \dfrac1{n^3}$$

Where $R$ is Rydberg constant, $c$ speed of light and $n$ is an integer greater than $0$.

I got this expression for freqency by replacing $v$ with expression for Bohr velocity and $r$ with expression for Bohr radius.

Actual emitted frequency is $$\nu^\prime = Rc \left(\dfrac{1}{(n-1)^2} - \dfrac 1{n^2}\right) = Rc \left( \dfrac{2n -1}{n^2(n-1)^2}\right)$$

Now if I let $n \to \infty$ in $\nu'$ I get $0$ not the expression for $\nu$.

When I asked someone, they said that for large $n$, $2n - 1 \approx 2n$ and $n - 1 \approx n$. So $$\nu' = Rc \left( \dfrac{2n -1}{n(n-1)^2}\right) = \dfrac{2nRc}{n^4} = \nu.$$

I think this is post-hoc way of reasoning. If I replace $n - 1 \approx n$ in $$\nu^\prime = Rc \left(\dfrac{1}{(n-1)^2} - \dfrac 1{n^2}\right)$$ I get $0$ but I do the same in $$\nu^\prime = Rc \left( \dfrac{2n -1}{n^2(n-1)^2}\right),$$ I get the correct answer.

Surely I am missing something or the statement is incorrect. Please help me.


This is maybe a bit of hidden terminology, but when we say things like

for large $n$

we do mean that the result is only approximate and that the approximation gets better and better the larger $n$ is. In this specific case, the result is asymptotic, which means that you need to be quantitative about the limit.

Thus, by doing the expansion correctly you can show that $$ \dfrac{1}{(n-1)^2} - \dfrac 1{n^2} =\dfrac{2n -1}{n^2(n-1)^2} =\frac{2}{n^3} + \frac{3}{n^4} + O(n^{-5}) $$ as $n\to\infty$, where the error term is expressed in big O notation. This result is independent of whether you use $\frac{1}{(n-1)^2} - \frac 1{n^2}$ or $\frac{2n -1}{n^2(n-1)^2}$, because they're equal, but if you take the limit too early in the first one then you'll only get $\lim_{n\to\infty}\frac{2}{n^3}=0$ on the right-hand side, which is the discrepancy you notice.

  • $\begingroup$ Oh ok so we taking first degree approximation of $\nu^\prime$, right ? $\endgroup$ – user8277998 Oct 2 '17 at 16:48
  • 2
    $\begingroup$ Yup, precisely, though it's more accurate to describe it as a leading-order approximation. $\endgroup$ – Emilio Pisanty Oct 2 '17 at 16:48
  • $\begingroup$ Thank you very much, this had me confused for the past day. $\endgroup$ – user8277998 Oct 2 '17 at 16:50

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