Solving for R in Atomic Spectroscopy | Get Homework Help and Tips

In summary, the student discusses their lab experiment using a spectrometer to find spectral values for Helium and using an excel spreadsheet to find the wavelengths for Hydrogen. They were tasked with solving for R, but all of their values had a 99.9% error. They question if they are calculating this wrong and realize their professor may have made a mistake in providing the correct value for R. After recalculating, they determine that the correct value for R should be 1.10 x 10^7 m^-1, not 1.10 x 10^-7 m^-1 as originally given by their professor.
  • #1
mainzelmadchen
5
0

Homework Statement


For our lab, we looked through the spectrometer to find the spectral values for Helium, which after being plugged into an excel spreadsheet, yielded an equation which I could use to find the wavelengths for Hydrogen. We were told to solve for R, which I did. However, at the end, all of my values have a 99.9% error! Even when I use my professor's wavelength values, I do not get the actual value of R = 1.10x10^-7 m-1. Am I calculating this wrong? My professor's wavelength values for #1,2,&3 respectively are 675, 450, & 415 nm.

Homework Equations


1/λ = R ((1/2^2)-(1/ninitial^2))


The Attempt at a Solution


1) 1/748.60606 nm = R ((1/2^2)-(1/3^2)); 0.0013358161 = R ((1/4)-(1/9)); 0.0013358161 = R (0.13888889); R = 0.0096178758 nm x (1 m/1 x 10^9 nm) = 9.6178758 x 10^-12 = 9.6 x 10^-12

2) 1/460.72727 nm = R ((1/2^2)-(1/4^2)); 0.0021704815 = R ((1/4)-(1/16)); 0.0021704815 = R (0.1875); R = 0.011575901 nm x (1 m/1 x 10^9 nm) = 1.1575901 x 10^-11 = 1.2 x 10^-11

3) 1/384.96970 nm= R ((1/2^2)-(1/6^2)); 0.0025976070 = R ((1/4)-(1/36)); 0.0025976070 = R (0.22222222); R = 0.011689232 nm x (1 m/1 x 10^9 nm) = 1.1689232 x 10^-11 = 1.2 x 10^-11
 
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  • #2
1/748.60606 nm
[...]
R = 0.0096178758 nm x (1 m/1 x 10^9 nm)
nm should be in the denominator in both cases. This would have been easy to spot if you did not drop the units in between.
It is the same for the other calculations.
 
  • #3
Ok, I see what you mean. At the end, I should have done (for number one) R = 0.0096178758 nm^-1 x (1 x 10^9 nm/1 m) which results in 9.6178759 x 10^6 m^-1. My answer is even more off. I tried this with my instructor's value to see if I get the correct answer. I did 1/675 nm = R ((1/2^2) - (1/3^2)) and got the result of 1.0666667 m^-1 x 10^7 and it should be 1.10x10^-7 m-1. Is there something else I am not doing correctly?
 
  • #4
I think I figured it out. I think my instructor made a mistake on this. I believe the answer should be 1.10 x 10^7 m^-1 and not 1.10 x 10^-7 m^-1. After calculating the normal equation constants, this is what I came up with.
1/wavelength = (2.18 x 10^-18 J / ((6.63 x 10^-34 J x s)(3.00 x 10^8 m/s)) x ((1/2^2)-(1/ninitial^2))

1/wavelength = (1.0960282 x 10^7 m^-1) x ((1/2^2)-(1/ninitial^2))
 
Last edited:
  • #5
As a simple cross-check, R should have the same order of magnitude as 1/λ.
1.10 x 10^7 m^(-1) looks good, 1.10 x 10^(-1) m^(-1) is wrong.
 

1. What is atomic spectroscopy?

Atomic spectroscopy is a scientific method used to analyze the composition and structure of atoms and molecules. It involves the use of instruments to measure the interaction between electromagnetic radiation and atoms, which can provide information about the energy levels and electronic configurations of the atoms being studied.

2. What are the different types of atomic spectroscopy?

There are three main types of atomic spectroscopy: atomic absorption spectroscopy, atomic emission spectroscopy, and atomic fluorescence spectroscopy. These techniques differ in the way they measure the interaction between atoms and radiation, and each has its own advantages and limitations.

3. What can atomic spectroscopy be used for?

Atomic spectroscopy has a wide range of applications in various fields, including chemistry, physics, astronomy, and environmental science. It can be used to identify and quantify elements in a sample, determine the chemical composition of unknown substances, and study the behavior of atoms and molecules in different environments.

4. How does atomic spectroscopy work?

Atomic spectroscopy works by passing a beam of electromagnetic radiation (such as light) through a sample containing atoms. The atoms absorb certain wavelengths of the radiation, causing them to become excited and emit light of their own. By measuring the amount and type of light emitted, scientists can determine the elements present in the sample and their concentrations.

5. What are the advantages of using atomic spectroscopy?

One of the main advantages of atomic spectroscopy is its high sensitivity and specificity. It can detect and quantify trace amounts of elements in a sample, making it a valuable tool in many scientific fields. Additionally, it is a relatively fast and non-destructive technique, allowing for the analysis of a wide range of samples without damaging them.

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