What is the final expression we get for wave number in a hydrogen emission spectrum? Let us say for example from 6 to an orbit $n$. Considering that it is an emission spectrum, I think it would be right to say $n<6$. And wave number is a positive quantity so I thought it would be $R_h(\frac{1}{n^2}-\frac{1}{36})$ and slope of the graph between wave number and $\frac{1}{n^2}$ would be $R_h$. But I would like to ask if it is right though. I recently wrote an exam who says the slope is $-R_h$, where $R_h$ is Rydberg's constant.



Firstly, the question doesn't seem logical as n can take only 5 discrete values here, i.e. 1,2,3,4,5. So the graph will have just 5 isolated points. Still, if you do have to plot a continuous linear graph through these points, the slope of the line will be positive and equal to +Rh. So you are right.

  • $\begingroup$ There are no restrictions on how high $n$ can be. It can also be infinite and it's totally fine. Also, plotting continuous line across discrete data points is just wrong (you can estimate the slope though). $\endgroup$
    – andselisk
    Jan 16 '19 at 20:13
  • $\begingroup$ Thanks Ankit. So it should increase with increase in (1/n^2) then. Thought so too. The key is probably wrong then, thanks! $\endgroup$
    – Apoorv Pal
    Jan 17 '19 at 1:23
  • $\begingroup$ @andselisk, the question talks about de-excitation from the 6th shell, so I stated that n can take values only from 1 to 5. Thus, 5 discrete points. $\endgroup$ Jan 17 '19 at 8:05

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