It may be helpful to review some right triangle trigonometry. Given the right triangle: C = 90º


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1 Ryn Lenet Pge 1 Chemistry 511 Experiment: The Hydrogen Emission Spetrum Introdution When we view white light through diffrtion grting, we n see ll of the omponents of the visible spetr. (ROYGBIV) The diffrtion (bending) of light by diffrtion grting will seprte the eletromgneti rdition s funtion of its wvelength. In ontrst to white light soure, when we look t soure of eletromgneti rdition from n exited gseous element we do not see ontinuous spetrum; we only see speifi bright lines of olor. This phenomenon ultimtely helped to unrvel some of the mysteries of tomi struture. In this experiment we will view the bright line spetrum for the simplest element, hydrogen. We will mke some simple mesurements nd then pply the eqution developed by the fther nd son tem of Sir W.H. Brgg nd Sir W.L. Brgg to lulte the wvelengths of light viewed. Biogrphil note: In 1913 the Brggs determined tht the sttering of X rys (short wvelength eletromgneti rdition) by rystls ws funtion of the wvelength of the inident Xry nd the distne between the lyers of the rystl. By using only speifi wvelengths of eletromgneti rdition, they were ble to determine the struture of seleted rystlline solids. For their work, the Brggs were wrded the Nobel Prize in Physis in Brggs Lw is the eqution: n = d sin The business prt of the eqution is = wvelength, d = distne between the lyers of the rystl nd = the ngle of diffrtion. [n is tully whole number whih n be 1,, 3, termed the order of diffrtion. You will be observing 1st order diffrtion, so, n will be equl to one (1) for us.] This eqution, originlly developed for the determintion of rystl struture, hs been pplied in the quest to determine the struture of mny other sttes of mtter. Bkground It my be helpful to review some right tringle trigonometry. Given the right tringle: b B C = 90º A Angle C is the right ngle, side is opposite of C nd is lled the hypotenuse. The other two ngles, A nd B hve opposite sides nd b respetively. 1
2 Experiment: The Hydrogen Spetrum Pge The following reltionships pply: Pythgoren Theorem: + b =. Sines: sin A = /, sin B = b/. Cosines: os A = b/, os B = /. Tngents: tn A = /b, tn B = b/. (dpted from essed 1/07/06) We n pply the ides presented bove to our experimentl design. Power Soure Gs Tube Spetrl Lines Distne(s) b Distne Diffrtion Grting In this experiment you will look through the diffrtion grting to view the hydrogen spetrum produed by the high voltge soure nd the gs tube. You will mesure distne, the distne from the grting to the soure nd multiple distnes long b, the distne from the soure to eh spetrl line observed. If we look t the mesurements tht re mde, we see tht we hve right tringle: b
3 Pge 3 Experiment: The Hydrogen Spetrum We will dpt Brggs Lw to llow us to solve for eh wvelength of light viewed using the eqution: = d sin() Where = the wvelength of light, d = the sping between the lines of the diffrtion grting nd = the ngle of diffrtion. Our next tsk is to determine how to rrive t sin() from the experimentl dt. The sine funtion of n ngle is the length of the opposite side divided by the length of the hypotenuse. For ngle we know the length of the opposite side (b) but we hve not mesured the length of the hypotenuse. We do hve mesurement for the other side of our right tringle, side. Applying the Pythgoren Theorem, + b = we will be ble to ompute the length of the hypotenuse nd then lulte the sine of the ngle. + b = ( + b ) = Our working eqution to lulte eh spetrl line observed my be derived. = d sin b sin = b = d = d b + b Proedure: 1. The lb will be set up ording to the digrm on pge.. In drkened room, one student will look through the diffrtion grting to view the spetrl lines of hydrogen. 3. A seond student will move penil down distne b until the penil nd spetrl line of hydrogen re ligned. This distne will be reorded long with the olor of the line observed. 4. This mesurement will be repeted for eh of line spetrl lines observed. 3
4 Pge 4 Experiment: The Hydrogen Spetrum Worked exmple: = 35.0 m ; b = 17.0 m ; d = 1000 nm = d b + b = = nm Other useful formul nd onstnts: = f = 3.00x10 E = hf E= h 10 h = 6.66x10 m s 34 Js photon Resoures: essed 1/9/05 essed 1/9/05 essed 1/07/06 essed 1/9/05 4
5 Pge 5 Nme: Ryn Lenet Chemistry 511 The Hydrogen Spetrum Lb Report [Rell tht d represents the distne between lines in the diffrtion grting. If you look t you grting your will red the following listing: 500 lines/mm, i.e., eh millimeter (mm) ontins 500 lines. So, the inverse of this, i.e., 1/500 should be the number of millimeters between line nd n djent line. Thus, (1/500) mm/line. The result should be: d = 0.00 mm between lines or 000 nm between lines.] Experimentl Dt nd Clultions: Distne Distnes b Color Observed 90 m 90 m 90 m Clultion Formul 0.1 m 3.3 m 3.0 m Violet Aqu/Blue Red Wvelength x 10^5 m x 10^5 m 6.7 x 10^5 m Lmbd= d(b/squreroot(^+b^) Energy x 10^19 J/photon 3.96 x 10^19 J/photon.96 x 10^19 J/photon E= hf E=h(/lmbd) Show representtive wvelength nd energy lultions. Then fill in the tble with you results. 1. The energy vlues lulted represent the energy emitted by the eletron when it trnsitions from higher energy level to lower. The red line in the hydrogen spetrum is the result of trnsition from the n=3 to the n= energy levels. Wht energy level trnsitions re represented by the other lines in the spetrum? The energy level trnsitions represented by the other lines in the spetrum re Hbet, whih is trnsition from level 4 to the seond level tht yields qu/blue line nd Hgmm whih is trnsition from level 5 to the seond level tht yields violet line.. Using your results, explin how E, the differene in energy between levels, vries s n, the energy level, inreses. 5
6 Pge 6 When n gets lrger, the lines strt getting relly lose together. As n gets lrger, 1/n^ gets smller, so there is less nd less differene between onseutive lines. The series hs limit. As n gets lrger nd lrger, the wvelength gets loser nd loser to one prtiulr vlue. 3. The spetrum you observed is the visible spetrum for hydrogen, lso lled the Blmer series. There re two other series for hydrogen: the Pshen nd the Lymn series. The Pshen series ours t wvelengths longer thn those observed in the Blmer series while the Lymn series ours t wvelengths shorter thn those observed in the Blmer series. In whih prt of the eletromgneti spetrum is eh of these spetrl series? Plese explin your resoning. Rell tht the lines in Blmer series n be represented s eletron energy level hnges (trnsitions) from nlevels > (i.e., 3, 4, 5, 6, ) down to n f =. So, for the Blmer series of lines, some of the eletron energy level trnsitions re: n i = 3 > n f = ; n i = 4 > n f = ; n i = 5 > n f = ; et. One of these new spetrl series (Pshen or Lymn) involves trnsitions from higher nlevels to n f =1 nd the other involves trnsitions from higher nlevels to n f = 3. Whih eletron trnsitions in the hydrogen tom produe eh series? Is this onsistent with your nswer to #? Plese explin your resoning. A pttern of spetrl lines, either bsorbed or emitted, re produed by the hydrogen tom. The vrious series of lines re nmed ording to the lowest energy level involved in the trnsition tht gives rise to the lines. The Lymn series involves jumps to or from the ground stte (n=1). It lies in the ultrviolet with series limit t 91A. The Blmer series(in whih ll the lines re in visible region) orrespond to n=. The Pshen series orrespond to n=3. This series re the shortwve infrred lines. The spetrum of rdition emitted by hydrogen is nonontinuous. Therefore the lines seen in the imge re the wvelengths orresponding to n= to n= infinity. This is onsistent with my results in question. 6
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