Atomic Emission Spectra PDF

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Summary

This is a lab write up for the analysis of an atomic emission spectrum. This is from the lab section of chem 200 or chem 202. ...

Description

Riley McConaughey! Chem 202 - 1! 821832161! Lab Partner: Jonathan Davani!

Chem 202 - Section 1, Experiment 8! Atomic Emission Spectra ! Procedure: For this laboratory experiment most procedures were done to specification, except for part C. In part C, the procedure said to spray the ion onto the flame, 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

Riley McConaughey! Chem 202 - 1! 821832161! Lab Partner: Jonathan Davani! 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

502

366

588

3593

668

946

707

910

434

1329

485

4095

588

2281

657

4095

Hydrogen

Helium

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

Riley McConaughey! Chem 202 - 1! 821832161! Lab Partner: Jonathan Davani! Sample Calculations (Part B) The Rydberg Equation: 1 = -1.0966776 x 107 m-1( 1 " " " λ" " " " n12

1 )! 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)" " √ 1.0966776 x 107 m-1(502) - (22)" a! n2 = | 1.0966776 x 107 m-1(22)(588)" " √ 1.0966776 x 107 m-1(588) - (22)" 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 flame, similar to scarlet red

Sodium Chloride

Yellowish orange flame

Copper Chloride

Light green flame

Lithium Chloride

Hot pink, or scarlet red flame.

Calcium Chloride

Dark and bold orange flame

Barium Chloride

Yellow flame

Strontium Nitrate

Reddish orange flame

Riley McConaughey! Chem 202 - 1! 821832161! Lab Partner: Jonathan Davani! Barium Nitrate

Yellow flame

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 specific 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 specific elements from their spectrum. ! 6. One ion was seen and the other ion was not seen during the flame test because the flame 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 flame color is a characteristic property from our observations, this is why the different ions have different colors....