1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ your original statement is wrong $\endgroup$ – The Imp Dec 27 '15 at 18:14
2
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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

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.