The neon bulb was red and the bright line spectra began with many red lines followed by red-orange lines which was followed by smaller amount of yellow lines. Then, there was a gap afterwhich came green lines followed by blue lines. The last light source we tested, argon, had a purple bulb with a line spectra that appeared to be solely made up of violet lines. During the second part of the experiment, we vaporized each metallic ion in flames and collected the resulting color of the flame. With this information, we consulted the chart that shows the representative wavelength and wavelength region in nanometers of different colors to make an educated guess on the possible wavelength of each collected flame color. For the lithium ion we observed a red colored flame (690 nm wavelength), the sodium ion had a yellow orange flame (597 nm), the potassium ion had a salmon pink flame (420 nm), the calcium ion a light orange flame (607 nm), the strontium ion a blood orange flame (678 nm), the barium ion a light green flame (550 nm), the copper ion a green flame flame (593 nm), and the unknown ion with a red flame (690 nm) very similar to that of the lithium …show more content…
Using the formula 풱= cλ in which c was equivalent to 3.00x 108m/s, we calculated the frequency of each emission. The lithium ion had a frequency of 4.35 x 1014Hz, the sodium ion, 5.03 x 1014Hz, the potassium ion, 7.14 x 1014Hz, the calcium ion, 4.94 x 1014Hz, the strontium ion, 4.42 x 1014Hz, the barium ion, 5.45 x 1014Hz, the copper ion, 5.06 x 1014Hz, and the unknown, 4.35 x 1014Hz. Then we used the formula E=hx풱 where h equals 6.629 x 10-34J/s to find the energy of each emison. The result for the lithium ion was 2.88 x 10-19J , for the sodium ion, 3.33 x 10-19J, the potassium ion, 4.73 x 10-19J, the calcium ion, 3.27 x 10-19J, the strontium ion, 2.93 x 10-19J, the barium ion, 3.61 x 10 -19J, the copper ion,3.35 x 10-19J, and for the unknown, 2.88 x 10-19J. As can be seen from the above results, we concluded that the unknown substance was the lithium ion because of its distinct red color emission during the flame