Kayley KilcoynePhysics 132Guy RosinMVAtomic SpectraAbstract: In this experiment we looked at the light emitted by different elements such as sodium, mercury and hydrogen. We also looked at continuous light. We measured the distance of the colored lines to find the wav lengths that were present in that specific element. By finding the wavelength of the different kinds of light we could then find the frequency.Questions:1.Hydrogen (H) b = 1.14Line Color a_left (m) a_right (m)a_average(m)sin Red 0.235 0.23 0.2325 0.199 6.567E-07Green-Blue 0.19 0.19 0.19 0.164 5.412E-07Indigo 0.18 0.175 0.1775 0.154 5.082E-07Violet? 0.15 0.16 0.155 0.135 4.455E-07I found the wavelength calculations by using the formula λ=dsinθm. Next I used the equation that 1average2−1b2dsinθ=.01097 ¿) . The d value was 3.33E-6, and this was found on the lens we look through to see the light. 2.Continuous b = 1.14Color a_left (m) a_right (m)a_average(m)sin Red 0.24 0.22 0.23 0.1977 6.5241E-07Violet 0.162 0.17 0.166 0.144 4.752E-073. The wavelengths of green light in the Mercury sample had a wavelength of 5.1E-7. The wavelength of yellow light in the sodium sample had a wavelength of 6.18E-74. The frequencies of violate and red color light is found by the equationc=λv. C is the speed of light 3E8 m/s. Take the wavelength found for red which is 6.5E-7 and then put it into the equation results in the frequency of red light equaling 4.61E-14 then you have to convert this and the frequency is461 THz for red light and then you fill in the wavelength of violet light for 4.75E-7 into the same equation and the frequency for violet light is 632 THz The frequency for violet light would be higher than the frequency for red light. 5. Using the equation c=λv, to find the frequency of light, you have to fill in the wavelength for the λ . For example in hydrogen the red light has a wavelength of 6.56E-7 and the you divide c which is 3E-8 by this number and you get 4.51E-14 then you have to convert this to nm which would be 451 THz for red light. Using this same equation the frequency of green-blue light would be 545 THz, indigo light would be 590 THz and violet light would be 674 THz. Blamers formula is 1λ=.01097(122−1n2). The expected value for frequency of red light should be 4.57E14, green-blue should be 6.17E14 and indigo should be 6.91E14. I think the numbers I got were fairly close to the expected values. I found these expected values by knowing that the n value for red was 3 , green-blue was 4 and indigo was 5. Then I plugged these n values into the blamers formula and got the expected values out.6. In this lab we were looking at emission spectra because emission spectra is the amount of light that is emitted at different wavelengths. The absorption spectra is the white light between the colors, this is exactly the opposite of emission spectra. The absorption spectra would have looked like a ray of continuous colors with dark line missing and the emission spectra is the opposite with colored lines. Emission spectra shows clearly distinct lines. 7. If the transition of the electron was from n=4 to n= 3 then the wavelength would not be visible. The wavelength would be found by using the equation .01097(1/3^2)-(1/4^2)= which would equal 5.33E-4. Then to find the wavelength you would do 1/5.33E-4 and you would get a wavelength of 1875nm. This wavelength is much too high to be seen by the human eye. The visible light spectrum ranges from 400-800nm. 8. A star that has emissions with wavelengths of 486.1, 546.0, 589.0 and 656.3. Wavelength of 486.1 would be a blue color, 546.0 would be a green color, 589 yellow color and 656 would be a red color. This tells us that he star is made out of sodium. Hydrogen and Mercury both give off violet light, which is not present in this example, so this star is made of sodium.Conclusion: There was a lot of room for human error in this experiment because you had to follow the lines with your eyes in the dark and it was difficult to tell where the lines started. I think that the values I got for the wavelengths were very close to the expected values of the wavelengths found by using the blamer’s formula.We were able use the data that we found from the wavelengths colors to determine what element we were looking
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