Atomic Emission Spectra

Atomic Emission Spectra ! Procedure: For this laboratory experiment most procedures were done to speciÞcation, except for part C. In part C, the procedure said to spray the ion onto the ßame, however we dipped a long toothpick into the ion and burned the saturated toothpick instead.! Experimental Data (Part A) Light Being Studied Color Overhead Fluorescent Hydrogen Helium Calculated Results (Part A) Observed Wavelength (nm) Intensity of Color Red 610 Intense Orange 590 Medium Yellow 570 - 580 Weak Green 530 - 570 Intense Blue 480 - 520 Medium Violet 410 - 480 Medium Red 650 Medium Orange None None Yellow 585 Weak Green None None Blue 480 Medium Violet 430 Weak Red 650 Intense Orange 600 Medium Yellow 580 - 590 Intense Green None None Blue 440 Weak Violet 300 Weak The Results for this part of the lab will not be calculated.! Sample Calculations (Part A) The Results for this part of the lab will not be calculated.! Experimental Data (Part B) Light Being Studied Wavelenth From Ocean Spectrometer (nm) Intensity (Counts) 389 322 448 259 Hydrogen Helium 502 366 588 3593 668 946 707 910 434 1329 485 4095 588 2281 657 4095 Calculated Results (Part B) Hydrogen Value Used for R ! Wavelength N1 For Hydrogen (2) 389 8 448 5 502 1.096776 x 107 N2 (Calculated) 588 D! 2 4 3 668 3 707 3 Sample Calculations (Part B) The Rydberg Equation: 1 = -1.0966776 x 107 m-1( 1 - 1 )! " " " λ" " " " n 12 n22! Because we are using hydrogen, all n1 values are 2! The derived Rydberg equation to solve for n2 can be written as: ! a! n2 = | 1.0966776 x 107 m-1(n12)(λ)" " √ 1.0966776 x 107 m-1(λ) - (n12)" ! " " " a! λ = 389" " n2 = | 1.0966776 x 107 m-1(22)(389)" " √ 1.0966776 x 107 m-1(389) - (22)" a! n2 = | 1.0966776 x 107 m-1(22)(448)" " √ 1.0966776 x 107 m-1(448) - (22)" a! n2 = | 1.0966776 x 107 m-1(22)(502)" " 7 -1 2 √ 1.0966776 x 10 m (502) - (2 )" a! n2 = | 1.0966776 x 107 m-1(22)(588)" " 7 -1 2 √ 1.0966776 x 10 m (588) - (2 )" a! n2 = | 1.0966776 x 107 m-1(22)(668)" " √ 1.0966776 x 107 m-1(668) - (22)" a! n2 = | 1.0966776 x 107 m-1(22)(707)" " √ 1.0966776 x 107 m-1(707) - (22)" " " ! " λ = 448" " " λ = 502" " " λ = 588" " " λ = 668" " " λ = 707" " " " " " " " ! " " " " " n2 = 7.97043825849129 ≈ 8! " " " " " " n2 = 4.632799439256865 ≈ 5! " " " " " " n2 = 3.8215652097452804 ≈ 4! " " " " " " n2 = 3.2440423892611236 ≈ 3! " " " " " " n2 = 2.9671790790519488 ≈ 3! " " " " " " n2 = 2.8735167835620783 ≈ 3" " " Experimental Data (Part C) Solution Tested Observations (Flame Color) Strontium Chloride Reddish orange ßame, similar to scarlet red Sodium Chloride Yellowish orange ßame Copper Chloride Light green ßame Lithium Chloride Hot pink, or scarlet red ßame. Calcium Chloride Dark and bold orange ßame Barium Chloride Yellow ßame Strontium Nitrate Reddish orange ßame Barium Nitrate Yellow ßame Calculated Results (Part C) The Results for this part of the lab will not be calculated.! Sample Calculations (Part C) The Results for this part of the lab will not be calculated.! Discussion Questions: 1. The Rydberg equation works for hydrogen, but not helium because the Z value, or the number of protons in hydrogen is 1. This will not affect the outcome when we square it because 1 squared is just 1. Helium has a Z value of 2, this will alter the equation we have because 2 squared is 4, hence this is why the Rydberg equation will work for hydrogen, but not helium.! 2. Most of the wavelengths observed for hydrogen match the transitions. Some of the quantum numbers are slightly off and in the middle of the two energy levels. The origin for this error could be because of errors in measuring wavelengths of the light with the spectrometer.! 3. Because hydrogen only has one electron, the spectrum for helium is more complicated. The one electron causes less excited states, therefore making it a simpler spectrum. ! 4. Yes, the values from the ocean optics spectrometer can be used to assign speciÞc wavelengths to the transitions that we observed with the STAR spectrometer. However, they will not be exact, but pretty close.! 5. Because the wavelengths are based off of the spectrum, this means that scientists can determine speciÞc elements from their spectrum. ! 6. One ion was seen and the other ion was not seen during the ßame test because the ßame causes the element to become neutral. The electron is being dropped from a higher energy level when the light is produced. Light is not produced when the energy level goes up. This is why only one ion was seen. One of the ions is dropping energy levels, while the other ion is gaining energy. The ßame color is a characteristic property from our observations, this is why the different ions have different colors.