-1
$\begingroup$

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.

Thanks!

$\endgroup$
-1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 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 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. $\endgroup$ – Ankit Kumar Misra Jan 17 at 8:05

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.