1
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.