In case of single isolated atom if electron makes transition from nth state to the ground state then maximum number of spectral lines observed $ = ( n — 1)$.

Is the above statement true? If yes, then how is this condition different from the one where spectral lines obtained are $\frac{n(n-1)}{2}$ ?

NOTE- I know how the formula for latter came.

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    $\begingroup$ your original statement is wrong $\endgroup$ – The Imp Dec 27 '15 at 18:14

I think you are mixing two different $n$. If an atom has $N = \text{number of levels}$ then the number of transitions and therefore number of spectral lines is $N-1$. However, for energy levels in an atom it is common to use three numbers to label each energy levels. This numbers are $n,l,m$. Each quantum number can take different values:

  • $n$ can take any integer greater than 0.
  • $l$ takes values from 0 to $n-1$.
  • $m$ takes values from $-l$ to $l$.

As you see all the conditions depend on $n$. If you do the math, for a given quantum number $n$ you have $n^2$ levels. The total number of transitions from level $(n_1,l_1,m_1)$ to the other levels will be

  • $\sum_{n<n_1}^{n_1} n^2 = N=\text{number of levels}$

Therefore, the number of transitions is $N-1$. As you see you should not mix $N = \text{total number of levels}$ with $n$ quantum number.

Note: I did not take into account forbidden transitions.

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    $\begingroup$ This is a fairly inaccurate answer. Look at this and follow the links. hyperphysics.phy-astr.gsu.edu/hbase/quantum/helium.html $\endgroup$ – Gert Dec 27 '15 at 18:33
  • $\begingroup$ @Gert - I think the original intent was to describe a system which receives no excitation once emission starts. In that case, all emissions correspond to transitions from an energy level to a lower one, and your linked diagrams do not apply. Javier may want to clarify this. $\endgroup$ – WhatRoughBeast Dec 28 '15 at 3:26
  • $\begingroup$ @Gert - My explanation is a simple one. If you have a system with $N$ energy levels and you are at level $N$, then you have only $N-1$ decay transitions. For an Hydrogen atom this holds if all transitions are allowed. The important point is to distinguish between the quantum number $n$ and the total number of energy levels in the atom $N$. Each $n$ has $n^2$ levels. So if you are at level $(n_1,l_1,m_1)$ you can decay to $n^2$ for each $n<n_1$ (assuming you are in the highest energy state for that $n$). $\endgroup$ – Javier Puertas Dec 28 '15 at 12:38

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