Question #d0cfc

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Question #d0cfc

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Answer 1 · 2016-06-29T22:23:07

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 $R^{2}$ $"R"^2$ value" as asked and also "display equation" which returns the equation of the straight line.

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

This means the correlation coefficient $= \sqrt{0.8747} = 0.935$ $=sqrt(0.8747)=0.935$ which is very close to an exact correlation of $1$ $1$ - 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:

$d = 33.1 cm$ $d=33.1"cm"$ and $λ = 659.4 nm$ $lambda=659.4"nm"$

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

$\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$ $1/lambda=R[1/n_1^2-1/n_2^2]$

$R = 1.0974 \times 10^{7} m^{- 1}$ $R=1.0974xx10^(7)" ""m"^(-1)$

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

www.daviddarling.info

The red emission line will be the lowest in energy so will have the longest wavelength corresponding to $n = 3 \to 2$ $n=3rarr2$ .

Putting this into the Rydberg Expression:

$\frac{1}{λ} = 1.0974 \times 10^{7} [\frac{1}{2^{2}} - \frac{1}{3^{2}}]$ $1/lambda=1.0974xx10^(7)[1/2^2-1/3^2]$

(You made a mistake in your annotation and wrote $\frac{1}{2}$ $1/2$ instead of $\frac{1}{2^{2}}$ $1/2^2$ )

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

$:.lambda=656.1" ""nm"$

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.

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Question #d0cfc

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