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Suppose you have a collection of large amount of Hydrogen atoms in $n$th state($n-1$th excited state). They have to go to their ground state($n$=1).

Going from $n_1$ to $n_2$ makes a unique spectral line. An atom cannot raise its $n$, it can only decrease it.What is the minimum number of $H$ atoms needed to view all spectral lines? Although changing of $n$ is random, we may assume that we are lucky.e.g. for $n=3$ :

We have to see $$3 \to 1,2\to 1, 3\to 2$$ Minimum number of atoms is $2$. One will go to $3 \to 2 \to 1$ and other will $3\to 1$.

When $n=4$,

We have to see $4\to 3,3\to 2,2\to 1,4\to 1, 4\to 2,3\to 1$ Minimum number is $4$ :

$$4\to 1, 4 \to 3 \to 1,4\to 2\to 1,4\to3\to2\to1$$

How can we generalize it for any $n$? Please add appropriate tags.

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2 Answers 2

up vote 1 down vote accepted

Consider the transitions $a\to b$, where $a>\frac{n}{2}$ and $b\le\frac{n}{2}$. The number of such transitions is $$\Bigl\lceil\frac{n}{2}\Bigr\rceil\Bigl\lfloor\frac{n}{2}\Bigr\rfloor\ .$$ Moreover, these transitions are mutually exclusive in the sense that a single decaying electron can only include at most one of these transitions, for if we have a decay $$a\to b\to\cdots\to a'\to b'$$ then $$\frac{n}{2}\ge b\ge a'>\frac{n}{2}$$ which is impossible. So at least $$\Bigl\lceil\frac{n}{2}\Bigr\rceil\Bigl\lfloor\frac{n}{2}\Bigr\rfloor$$ atoms are necessary. But by considering the cases of $n$ even and odd separately we find that this is equal to $$(n-1)+(n-3)+(n-5)+\cdots\ ,$$ and @Hagen von Eitzen shows in his answer that this number is sufficient.

To sum up, you cannot do better, and need not do worse, than this number of atoms.

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You need $n-1$ electrons starting from the top level to perform $n\to n-1,\ldots n\to 1, n\to 1$. After that you have one electron available at all lower levels. You need $n-2$ electrins on level $n-1$, but already have one, i.e. you need $n-3$ "new" electrons to perform $n-1\to n-2,\ldots n-1\to 2,n-1\to 1$. This goes on, i.e. in total we need $$(n-1)+(n-3)+(n-5)+\ldots +(n-(2m-1)) $$ electrons where the number $m$ of summands is such that $2m-1\le n<2m+1$ (or $2m-1<n\le 2m+1$, it doesn't matter). The sum of the first $m$ odd numbers is $m^2$, so that the total count is $mn-m^2$ with $m=\lfloor n/2\rfloor$ (or $m=\lceil n/2\rceil$).

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Does this imply that if one wishes for every electron to reach the ground state and also to have every possible path taken, some paths will have to be taken more than once? –  Just_a_fool Apr 30 at 20:19

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