Error equation for the balmer series

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SUMMARY

The discussion focuses on calculating the error in the wavelength (λ) determined by the Bohr model's Balmer series. The participant uses the equation \(\frac{1}{\lambda} = R \left(\frac{1}{n^2_f} - \frac{1}{n^2_i}\right)\), where R is the Rydberg constant with an error of ±1 in the last digit. The confusion arises from the presence of λ in the error equation, leading to a need for clarification on how to isolate λ and perform error analysis correctly. The final equation for error in λ is expressed as δλ = λ²(1/K)δR, where K represents the difference in the squared terms.

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Homework Statement


I have to find the error in λ determined by the Bohr model/Balmer series. I am a bit confused with this so I'd like someone to double check my work. I don't know the exact name of this method but basically all you do is take the differential of all the values which have an error to find the error of the determined value.


Homework Equations



\frac{1}{\lambda} = R (\frac{1}{n^2_f} - \frac{1}{n^2_i} )
nf and ni are discreet values so they have no error in them
R is the Rydberg constant and has an error of +-1 in the last digit so
R = (1.097 +- 1 ) * 107 m-1


The Attempt at a Solution


\frac{1}{\lambda} = R (\frac{1}{n^2_f} - \frac{1}{n^2_i} )
\frac{\delta \lambda}{\lambda^2} = \delta R (\frac{1}{n^2_f} - \frac{1}{n^2_i} )
\delta \lambda = \lambda^2 (\frac{1}{n^2_f} - \frac{1}{n^2_i} ) \delta R
The reason I am confused with this is that I am trying to find the error in λ but λ shows up on the right hand side of my error equation. Also did I get the error in R in the right decimal place?
 
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First: replace everything in brackets with 1/K and solve for lambda.
Now find the error on lambda - should clear up your confusion.
 
that works thanks
 
So is the equation supposed to look like as follows: δλ=λ2(1/K)δR? I am so confused.
 
@dab353: welcome to PF;
Take it step-by-step.
Start with the full equation.
Substitute the 1/K in for the brackets... that's just to make it clear.
Solve that equation for lambda.
Do the error analysis like normal.
 

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