User:Justin Roth Muehlmeyer/Notebook/307L Notebook/Balmer

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SJK 19:16, 18 December 2008 (EST)
19:16, 18 December 2008 (EST)
Also some comments in Der's notebook.
SJK 19:14, 18 December 2008 (EST)
19:14, 18 December 2008 (EST)
I do see the left versus right study, but what I don't see is mention that you accounted for backlash when taking your data?

Introduction

The gas tube is filled with a monoatomic gas, which has an electric discharge passing through it. This electric discharge excites the atoms of the gas through collisions with its electrons and the gas atoms are thus put into a higher energy state where their total energy is greater than it is in a normal atom. As it returns back to its normal state, it gives off that excess energy gained from the electric discharge and emits electromagnetic radiation. This electromagnetic radiation can be viewed using a prism, where we break up its spectrum into its component wavelengths.

Quantum mechanics tells us that the energy levels of these excited atoms are discrete, meaning that we will observe discrete wavelengths in the spectrum. Each substance has its own set of discrete wavelengths, a unique "signature" that is the basis of the art of spectroscopy.

In this case we will use Hyrdogren, which contains only one electron, and we will observe the part of its spectrum that falls within the range of visible light.

Balmer used the Hyrdogren spectrum to make an empirical formula that would represent the wavelengths of these spectral lines. The formula developed a relationship that was further developed by Ryderberg who found it easier to work with the reciprocal of the wavlengths. The Balmer formula became...

$\frac{1}{\lambda}=R(\frac{1}{2^{2}}-\frac{1}{n^{2}})$

Where n=3 for the first line (largest wavelength), n=4 for the next line, etc. These numbers are now termed the principal quantum number, and represent the discrete energy levels above ground state.

Rydberg's constant today is considered to be: $R=1.0967758\cdot 10^{7} m^{-1}$

Which we will attempt to approximate in this lab.

We will do so by plotting $\frac{1}{\lambda}$ VS. $\frac{1}{2^{2}}-\frac{1}{n^{2}}$

The slope of this line will give us Rydberg's constant. We will use the least squares method to find the slope.

Calibration of the Spectrometer

Inserting the mercury lamp we began our calibration.

Adjusting the wavelength to 546.1 nm, then rotating the prism, we saw our green light and put it smack dab in the center of our crosshairs. Rotating the screw drive to 435.8 we saw the violet light in our cross hairs, validating the requirement that violet does exist at that point in the spectrum for mercury. We have calibrated our apparatus.

Our calibration was perfect, there is no need for correction.

Hydrogen

We had to switch out the gas tube because we were not seeing all the lines at first. The second gas tube was a bit more intense, and it enabled us to see more lines.

1st Run:

• violet 435.2 nm
• cyan 488.0 nm
• yellow 586.0 nm
• red 664.2 nm

2nd Run:

• violet 434.6 nm
• cyan 487.8 nm
• yellow 586.0 nm
• red 664.0 nm

Deuterium

1st Run:

• deep violet 409.3 nm
• violet 432.5 nm
• cyan 484.0 nm
• yellow 579.5 nm
• red 653.5 nm

2nd Run:

• deep violet 409.0 nm
• violet 432.8 nm
• cyan 484.3
• yellow 579.1
• red 653.0

Deuterium (or "heavy hyrdrogen") is an isotope of Hydrogen. It contains a proton AND an nuetron, meaning that its nuclear mass is twice that of hydrogen.

The Balmer series equation for Hydrogen assumes the mass of the atomic nucleus to be infinitely large compared to the mass of its electron. However, in reality, this is not the case, and we must correct for the finite nuclear mass. This is done using the reduced mass, which takes the mass of the electron into account and creates a center of mass for the nucleus-electron system.

Deuterium's nuclear mass is doubled, meaning that yes, it will indeed have a different Rydberg value from Hydrogen, only because its spectrum is slightly different. We see that our Rydberg for deuterium is larger than the Rydberg for Hydrogen. You can see that this would be the case from the data. The spectral lines for Deuterium are shifted slightly to shorter wavelengths than that of Hydrogen.

Krypton

After checking out Boleszek's labe notebook, we found that Na lamps don't comprise part of the collection currently, so we chose Krypton to check resolving power.

violet lines:

eyeballed closest:

1. 1a: 444 nm
2. 2a: 448 nm

distance: 4 nm

1. 1b: 434.9 nm
2. 2b: 436.1 nm

distance: 1.2 nm

Our resolution then can get very close to 1 nm. This is certainly enough to see very small differences.

Left vs. Right... Right Wins

We want to check to see how our data changes based on the direction we turn the knob...

to left side of green bar, Krypton...

coming from right: 555.0 nm from left: 554.4 nm

to right side of orange bar, Krypton...

coming from right: 578.0 nm from left: 577.3 nm

to left side of a violet bar, Krypton...

coming from right: 431.1 nm from left: 431.0

Conclusion:

It seems like the accuracy of the screw gets worse with increased wavelength.SJK 18:55, 18 December 2008 (EST)
18:55, 18 December 2008 (EST)
Or rather, the backlash or "slop" gets relatively worse.