Question #d0cfc

1 Answer
Jun 29, 2016

Answer:

Here's how you could do it:

Explanation:

It seems you need to produce a calibration curve for the distance measured against the wavelength of the emission line using Excel.

I got this from your data:

enter image source here

I assumed cm but I guess it could be mm - you should put this on your table.

Point 28 is an outlier - you should repeat this if you can.

I then used the 3 lines from the hydrogen lamp to read off the values of the wavelength.

I can't read your worksheet too well, it seems you are asked for a linear plot.

To do this enter your data in a table and make a scatter chart. Click on a data point and "add trend line". You can select "linear" and it will give you the line of best fit through the points.

You can select "display #"R"^2# value" as asked and also "display equation" which returns the equation of the straight line.

The #"R"^2# value is the square of the correlation coefficient for which I got #0.8747#.

This means the correlation coefficient #=sqrt(0.8747)=0.935# which is very close to an exact correlation of #1# - though I chose a binomial trend line.

I have manually read the wavelengths off the chart, though you could use the equation given.

So for red:

#d=33.1"cm"# and #lambda=659.4"nm"#

This is an experimental value. You are asked to compare this with the calculated value using the Rydberg Equation:

#1/lambda=R[1/n_1^2-1/n_2^2]#

#R=1.0974xx10^(7)" ""m"^(-1)#

The Balmer Series, which you are dealing with, involves transitions down to the #n=2# energy level:

www.daviddarling.info

The red emission line will be the lowest in energy so will have the longest wavelength corresponding to #n=3rarr2# .

Putting this into the Rydberg Expression:

#1/lambda=1.0974xx10^(7)[1/2^2-1/3^2]#

(You made a mistake in your annotation and wrote #1/2# instead of #1/2^2# )

#:.1/lambda=0.1524xx10^(7)" ""m"^(-1)#

#:.lambda=656.1" ""nm"#

So we have:

Experimental: #659.4" ""nm"#

Theoretical: #656.1" ""nm"#

That's reasonably good. You can work out the difference as a percentage as asked for in your error calculation.

Then work out #4rarr2# and #5rarr2# in the same way.