Find Rydberg's Constant Using Graphical Method

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Sintered
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Homework Statement



Note: Questions is targeted at Balmer series only (ie, shell 6 --> 2 )

Given the Rydberg formula, 1[tex]/[/tex][tex]\lambda[/tex]=R[1[tex]/[/tex](nf)2-1[tex]/[/tex](ni)2], describe a way of using a graphical method to determine the value of R.

Where nf=final shell (should always be 2 for Balmer series)
ni= initial shell (any shell from 6-->3)
[tex]\lambda[/tex]=Wavelength
R=Rydberg's constant

Homework Equations



Umm, I gave it above.

The Attempt at a Solution



I'm not sure...I understand the formula, I've done questions relating on the use of the formula, I have a good knowledge of the Balmer Series (as in Shell 3 --> 2 is given H[tex]\alpha[/tex], then 4-2 beta, etc), I know the empirical value of Rydberg's constant, but sadly I am completely stumped on how to do this. I'm pretty sure that this is done using gradients, but do not know what represents the Y and X values on the graph. Help will be much appreciated.
 
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Hi sintered, welcome to PF.

The assumption here is that you have a bunch of wavelengths from the hydrogen emission spectrum. You also know the Rydberg formula, but you don't know the value of R.

Question 1: Can you match nf values with wavelengths?

Usually situations like this are approached by "linearizing" the equation, i.e. bringing the equation into the form

y = a x + b

and fitting a straight line through the data points. Look at the formula.

Question 2: Can you picture in your mind what could possibly y and x be?

Once you have that, it should be easy to see what a and b ought to be.
 
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Thanks for the welcome Kuru! I think I get it now =].

From the wavelengths provided the values of nf and ni can be deduced (probably from the provided info and knowledge of Balmer Series), and then if you let y=1/[tex]\lambda[/tex] and x=1/(nf)2-1/(ni)2, then the value of X can be obtained and plotted on the X axis (?) with the corresponding value of the inverse wavelength on the Y axis (?). Rydberg's constant can then be obtained from the gradient.

I'm not sure if this is what you were hinting but its kind of making sense now. I won't ask you to provide everything in clarity, as this will be one of the topics in an upcoming assessment =D.