Portage Chem 103 Lab 2 Notebook

Creating Your Lab Notebook PDF

All lab notebook submissions must be a single file and the notebook pages must be in order in that

file or you will be asked to resubmit. The information below will instruct you on how to create a single

PDF file using an iOS or Android device. You are also welcome to use a regular computer scanner. If

you have any questions please reach out to your instructor.

iOS

1. Take individual pictures of each lab notebook page in order using your iOS device.

2. Go to the Photos app.

3. Find the album where the photos are located.

4. Tap the “Select” button in the top right corner of the screen.

5. Tap all of the images of your lab notebook.

6. Tap the “Share” button (square with an up arrow) in the bottom left corner of the screen.

7. Tap “Print”.

8. You will be able to review the images at the bottom of the screen. Make sure they are in the

correct order or you will be asked to resubmit.

9. Pinch and zoom in on the preview thumbnail to turn everything into a PDF.

10. Tap the “Share” button (square with an up arrow) in the bottom left corner of the screen.

11. You will be able to “Save to Files” or e-mail it to yourself.

12. Upload PDF in Canvas where appropriate as your lab notebook submission.

Android

1. Go to your Google Drive on your Android device.

2. Tap the + (plus sign) button in bottom right hand corner of the screen.

3. You should then select the option to “Scan”

.

4. Take a photo of one page of your lab notebook.

5. Tap the √ (check mark).

6. Tap the + (plus) sign that is in a square box in the bottom left corner of the screen.

7. Take a photo of the next page of your lab notebook.

8. Repeat steps 5-7 until all photos are taken.

9. Tap the √ (check mark).

10. Tap “Next” in bottom right corner of your screen.

11. There should be a popup that says PDF being generated.

12. Record the name of your file and select “Save” to save the PDF to your Google Drive.

13. Upload PDF in Canvas where appropriate as your lab notebook submission.

Lab 2 Transcript:
welcome back this is your lab experiment number two flame tests of metal cations ions and emission spectra let’s get started

by the end of this experiment you will have performed flame tests of metal cations ions to observe their characteristic colors you will observe and understand the emission spectra of atoms using gas discharge tubes and you will also rehearse calculating the energy of the Balmer series spectral lines that come from the bore model of the atom some concepts that we will review from the lecture include ground state electron configurations for both atoms and ions properties of light and the bore model of the atom we will review the three symbols that contain three important pieces of information the first is the principal quantum number n this tells us the relative energy level of the orbital the next is the letter that indicates the orbital type S P D or F and finally the superscripted number tells us how many electrons are included in the subshell the examples in the top right hand corner 2p4 describe the p orbital in the second energy level which contains four electrons this slide shows a portion of the graphic found in module 2.4 for electron configurations of atoms and ions it represents a strategy that we can use to infer the order of filling within a ground state electron configuration if we take a look at the top we see our starting point and we move in a side winding fashion from left to right left to right so our order of filling will be 1 S 2 S 2 P 3 S 3 P 4 S 3D 4 P 5S etc it should be worth noting that if this method is not particularly helpful to you you can also infer the order of filling by using the periodic table

here we are reminded that when an ion is formed atoms are either losing or gaining electrons in this specific lab we will be dealing exclusively with cations these are metal atoms that lose their outer shell electrons to form a positively charged ion main group elements are the electrons in main group elements the electrons that were added last are removed first for transition metals we will see that the removal of electrons occurs from the highest occupied s orbital and we’ll review this in our rehearsal here you can see the properties of light the electromagnetic spectrum just right of center on this scale you will see the narrow band of electromagnetic spectrum that is visible light the unit for wavelength is given in meters and the order of magnitude is between * 10 -6 and * 10 -7 m we’ll talk more about the unit that we’ll be using in this experiment later however it is worth noting that immediately to the left of visible light is infrared radiation and immediately to the right is ultraviolet radiation in part two of this experiment we will call back to the infrared and ultraviolet portions of the electromagnetic spectrum here we’re taking a little bit closer look at the visible light spectrum and we are noting on this image that the wavelength provided at the bottom of this slide is given in units of nanometers and specifically we can see that the visible light range is from approximately 750 nanometers and this would show us a red color of light and moving towards the right we see the visible light portion of the spectrum ending around 400 nanometers with violet and in between we have our traditional uh colors of the rainbow so red orange yellow green blue indigo and violet the work of Neil’s Boris discussed in module 2.2 and on this slide I simply want to refresh the phenomena of a ground state electron versus an excited state electron taking a look at our image on the slide we can see that as you move up the image the energy increases and so we will begin with our red sphere representing an electron and in this image we can see that energy in the form of a squiggly line is being added to that electron we’re going to add energy to electrons twice in this experiment in the first part we’ll be adding energy in the form of a flame in the second part we will be adding energy in the form of electrical energy once we do this the electron will move towards an excited state so it goes from lower energy to higher energy as a result of absorbing a particular amount of energy after a period of time the electron will then move as is shown in part C from the higher state to the lower state and when it does this it gives off energy in the form of a photon and we can measure certain things about this photon such as its wavelength and frequency which are indicators of the relative difference in the energy between the ground state and the excited state now on this slide we will show some of the equations that we will be using in this experiment especially in part two as we attempt to calculate the Balmer series of the hydrogen atom we know that the energy is equal to plank’s constant time the frequency of a photon and using a slight manipulation we also know that energy is equal to plank’s constant time the speed of light divided by the wavelength of the photon we also know that the difference in energy between two subshells within the atom can be calculated by taking the energy of the final subshell and subtracting from it the energy of the initial subshell and so delta E is equal to the absolute difference between the energy of our final subshell minus the energy of our initial subshell and we can actually set this equation equal to our previous equation that was used to calculate the wavelength of a photon so if we know the wavelength of a photon we can infer the difference in energy between the subshells within the atom and we do so using the following equation the energy of a subshell is equal to the negative K which is a constant from our book it describes the discrete energies of the electron orbitals and is equal to a value of 2.179 * 108 jewels and we divide this by n^ 2 where n is an indicator of the number of the energy level so as you can see here n is always going to be a whole number our initial or our final now in our experiment we will begin by conducting the flame tests of our metal cations ions before we do this we will rehearse writing their ground state and ionic electron configurations and then we will observe the characteristic colors of six different metal solutions using these known values we will then attempt to identify two unknown solutions in part two of this experiment we’re going to be observing first the hydrogen spectral lines and then we will rehearse calculating the energy of these photons that we observe in the gas discharge tube finally we will conclude with our summary as well as additional discussion questions for your lab report at this point before we go and write the ground state electron configurations for both the atom and the ion I would recommend that you have ready access to a periodic table as well as the order of filling for the subshells within the atom we will begin with lithium which is atom number three and so as lithium has an atomic number of three we’re going to write its ground state configurations filling from the lowest energy subshell 1 s 2 and then following 1 s we have 2s 1 and this is for the atom lithium but in this experiment we’re going to be using the lithium cation ion which has a charge of plus one and so to represent this ground state electron configuration we’ll remove the outer s orbital electron and we end up with a configuration of 1 s2 let’s move on to sodium sodium on the periodic table has a symbol of Na and it has an atomic number of 11 we will move rather quickly through these after we do the first few and so sodium with its 11 electrons will fill from lowest energy to highest 1 s2 2 s2 2p6 and as you’ve rehearsed these in your practice within the modules in the problem sets I’m sure by this point you recognize that these first three orbitals always add up to a total number of 10 and so our final electron will go into the 3s1 orbital now for sodium the ion which has a charge of plus one we’re simply going to be removing the outer s electron for sodium plus one we’ll simply be removing the outer 3s electron that will leave us with 1 s2 2 s2 and 2p6 for potassium whose symbol is k and has an atomic number of 19 we will first write the electron configuration for the ground state atom this is 1 s2 2 s2 2 p6 3 s2 3 p6 as you can see up to this point in time we’ve placed 18 electrons and we have one more to add this will leave us with 4 s1 for potassium the ion whose charge is 1 plus we’ll simply be removing the outer 4s electron and this will leave us with a configuration of 1 s2 2 s2 2 p6 3 s2 3 p6 with this rehearsal I would now encourage you to go through and write the configurations for copper whose symbol is cu and whose atomic number is 29 for strontium whose symbol is SR and atomic number is equal to 38 and for berium whose symbol is EA and whose atomic number is 56 once you finish writing these electron configurations resume the video and we will begin collecting data for part one of the experiment this is the first element of part one lithium as you can see we’ve given you back the ground state configuration and the electron configuration for the ion lithium plus one and in a moment I’m going to insert a wooden slip that has been soaked in a lithium cation solution and I’m going to ask you to record your qualitative description that is your description of the flame using your sense of sight and words so if you’re going to say is particular color you would want to be more specific than just saying red uh because multiple samples may be the color red so try to be descriptive um you might indicate an object in your immediate local environment that is of that particular red or you might try to relate this color of red to something more broadly known such as Crayola crayon colors in our quantitative description we are going to align the color that we are observing on screen with a particular wavelength or range of wavelengths from the electromagnetic spectrum so let’s get started here with lithium

and right off the bat we do see that lithium has a red if not reddish orange color indicating that there are uh it’s not just one wavelength but probably few different red colored wavelength

we’ll discard this split before it combusts and we will do that one more time again this is the lithium cation solution

so on your lab notebook you’re going to want to record your qualitative description uh you might say it’s a light colored red a red with potentially a mixture of oranges maybe even a bit of a violet hue in it uh and the wavelengths you would probably indicate a uh mid-range red colored wavelength maybe between 670 690 nmters

let’s move on to the next sample now we have sodium whose ground state configuration for the atom and the ion is shown here as uh in the ionic form it’s 1 s2 2 s2 2p6 and in a moment we’re going to insert the sodium wood splint into the flame

and now we will observe the sodium flame

and somewhat uh drastically different than our previous example we can see that sodium appears to be a very bright orange color uh maybe a very distinctive orange uh very different than just say a wood fire and we’re going to do that one more time so that you can write both your qualitative and quantitative observation

now if we left the splint in here long enough eventually the solution would completely burn off and we would just be burning the wood splint that would make it very difficult to tell the difference between uh the metal cation in the solution versus just the mood split now let’s record our results here our qualitative results uh it was a definite orange color and we would approximate that wavelength range to be about 580 and 600 nanometers that would be recorded as our quantitative observation or description let’s move now to potassium potassium whose symbol is K has an ionic configuration of 1 S2 2 S2 2 P6 3 S2 3 P6

again what is happening is that we are exciting the electrons within potassium ion solution and when these electrons are relaxing they’re giving off distinct colors of light that represent changes in energy levels here is potassium ion

and again a pretty drastically different color here uh at some point it doesn’t appear to be very different than the top of the flame except that it has somewhat of a pink component uh you can tell here now that I’ve left it in the flame too long that orange color is because the splint is beginning to burn so we’ll do that again for potassium had a very pretty somewhat pink or lilac color

and again once you see that orange begin to appear that just means that we’re burning our split

so that is the potassium ion for our qualitative description you can use your own or use the ones that I’m providing i would call it a a lilac or even a faint pink with some hues of purple and we’re going to approximate the wavelength range to be between 730 and about 750 nanometers for this metal ion now we’re going to take a look at copper so you can compare your rehearsed ionic configuration versus the one that’s provided here if we fill the subshells for copper according to our proper filling order we end up with an outer configuration that ends in 4s2 and 3d9 if we use this electron configuration for copper we will remove because it is a transition metal the outer 4 s2 electrons and this will give us copper 2+ let’s burn this uh metal solution and see what color copper burns

and here is the copper split

and immediately you can see some very distinct green wavelengths coming through that solution but also some blue So it’s not it’s not universally green uh we do have green and blue combined and the longer you leave that splint in there the more oranges you’ll see appear so remember the oranges that we observed after a period of time are just us due to us burning the splint so here’s round two for the copper ion and again at the very beginning you’re going to see those really beautiful greens almost like the aurora borealis and over time you can see those blues begin to appear and you could properly assume that the greens and the blues are because of different transitions within the copper ion so qualitatively we would describe that solution as being green and green blue and we’re going to give this one a little bit of a a broader range of wavelength we’re going to say that it’s between about 490 and 540 nmters let’s move now to strontium strontium’s outer configuration ends in 5s2 is a main group element and so when we write the configuration for the ion we’re simply removing those two outer s electrons and our ionic configuration is shown here to end in 4p6 let’s now observe strontium color qualitatively and quantitatively and this is strontium

now right off the bat we can see a a red color emanating from the split

the tricky part will be to distinguish this red for strontium from the previous red that we observed

and as we see the orange begin to creep in again that is from the split so again we’ll observe strontium and we’ll try to be detailed in our qualitative observation you might even consider pausing the video and going back and observing our previous red metal cation to be able to draw an easier distinction and this is why it’s important not only use qualitative observations but also quantitative observations and this again is strong

welcome back this is your lab experiment number two flame tests of metal cations ions and emission spectra qualitative observation you might even consider pausing the video and going back and observing our previous red metal cation to be able to draw an easier distinction and this is why it’s important not only to use qualitative observations but also quantitative observations

and this again is strontium all

[Music]

right and we have expired all of our strontium

so qualitatively for strontium uh we might describe it as being a a softer red uh which could possibly indicate that there are more uh different wavelengths of red that we’re observing here to possibly include even some oranges uh and we’re going to give this one a really broad quantitative description of being between all wavelengths 600 and 700 nanometers so it’s it’s not as distinct of a red as the one that we observed previously for lithium now on to our last known sample which is barerium again barerium uh is a main group element whose ground state atomic configuration ends in 6s2 and when it becomes an ion it becomes uh barerium 2+ with an outer configuration of 5p6 this is barerium let’s observe it both qualitatively and quantitatively

and now we have barerium

so it would appear as though we have some some green yellow involved uh it’s definitely a distinctly different green than the one we observed previously with copper but this yellow

that we observed is uh definitely not a a distinct or just a isolated yellow it’s a yellow combined with a green wavelength

and this is round two for barium

so to qualitatively describe barerium or or barerium 2 plus we would definitely see a yellow with hues of green um possibly indicating a broader spectrum of wavelengths like we observed previously with strontium ium which appeared to be a mixture of reds and oranges uh we can definitely distinguish barium from copper uh in that uh even though they both had elements of green their ranges of wavelengths were likely different and we would approximate barerium’s uh wavelength range to be again broad and between around 540 and 575 nanometers so now that we’ve identified and observed six known solutions lithium

sodium potassium copper strontium and berium we’re now going to attempt to ident identify two unknown solutions so uh a flame test can be used uh in a in a loose qualitative way and approximative uh quantitative way to try to identify unknown solutions in the lab so if we stumble upon this solution uh we can simply uh saturate a wood splint in the solution and then use the known color ranges to attempt to identify it and that’s what we’re going to do right now so for this part of part one uh we’re going to begin with unknown A we’re going to observe it qualitatively and quantitatively and then I’m going to leave uh the inference of the likely identity to you so I’ll be providing the qualitative and quantitative descriptions you can uh also feel free to uh use your own descriptions uh but I would like for you then to attempt to identify which of the six known solutions we are looking at with unknown solution so let’s get started

so here we begin with solution A

so right off the bat we can see that solution A burns a uh burnt orange or a a you might describe it as an autumn orange

this is round two for unknown A and hopefully what you’re doing right now is you’re looking back at your lab notebook and trying to identify which other known solution that you identified to be similar to unknown A

so unknown A had a definitive orange presentation and we would approximate unknown A’s wavelengths to be about between 580 and 600 nmters so using either your qualitative and quantitative description uh you might want to pause the video at this point and attempt to identify which metal cation ion was present in unknown sample A now we’ll go to unknown B

here is unknown B

and unknown B has a uh somewhat purplish pink color

we’ll do that one more time

and again we observe that uh very uh distinct purple pink color any elements of orange here are just coming from the splint

and so qualitatively uh we might describe unknown B as having a lilac or pink color with an approximate wavelength range of 730 and 750 nanometers now that we’ve gone through and collected our data on unknowns A and B attempt to identify them here before moving on to part two

this is part two of the experiment previously we were observing the characteristic flame color that was a result of electron transitions within metal cations ions we excited these electrons using the flame and we observed usually only one color coming out of the flame and this one color was indicative of the color of the metal cation in this part of the experiment we are now going to see behind the lines that is even though the light in part one appeared to only be one color it is actually multiple different colors all occurring within the atom at one time after all atoms have with the exception of hydrogen more than one electron and so at any given moment the electrons are transitioning between all of the energy levels now we are going to be using a device called a spectroscope to observe a gas discharge tube inside of this tube is only one element so hydrogen helium and mercury and using the spectroscope we will use a small piece of defraction grading which will act similar to a prism uh in that it will allow us to see the multiple different colors of light that are combined to make the one color of light that we are able to see using only our sense of sight so as you might imagine uh tiny little droplets of water that are suspended in the atmosphere after a rain uh they defract light so the white light or the light that comes from the sun gets dracted and broken down into the colors of the rainbow in a hydrogen atom we can predict the colors that will be seen using the bore model equations from the module and we are going to rehearse uh these calculations before we observe the gas discharge tubes so let’s get started with those equations we’re going to begin by predicting three electron transitions within the hydrogen atom specifically we’re going to rehearse the transition from the fifth to the second energy level the fourth to the second energy level and the third to the second energy level together these transitions are referred to as the Balmer series it’s a specific set of electron transitions to the second energy level within the hydrogen atom that all happen to occur or release energy in the form of visible light there are obviously more transitions that happen within the hydrogen atom but their wavelengths fall outside of the visible light range most notably in the infrared or the ultraviolet range so here on the screen you can see that we’re going to first calculate the difference in energy that results from an electron that transitions from the fifth to the second energy level then we are going to use that known energy to attempt to calculate the wavelength of that photon first in meters and then in nanometers let’s get started to begin and as it was rehearsed within the the lecture you’re going to calculate the energy of the electron at its final state and you are going to subtract from that the energy of the electron at its initial state in this instance the initial state is n= 5 and the final state is n= 2 and so it’s very important when calculating your energies which are equal to the k / n^2 that we make sure that we match up our initial and our final correctly so our final energy is going to be equal to K which again is our constant which has a value of 2.179* 10^ the8th power and that is has a unit of jewels we’re going to divide that by n^2 for our final so 2 ^2 this is our final energy and then we will subtract from that the same equation but for our initial energy level so that will be K over 5^2 now it’s really important when you enter these calculations into your calculator that you wrap these individual energies in parentheses uh anytime we’re doing calculations that involve numbers written in scientific notation we want to make sure that we use appropriate parentheses so K / 2^ 2 minusK / 5th^ 2 and again you probably want to conduct these calculations separately and then enter the values in and when you use the exact values given from each energy you should obtain a change in energy of -4.5759 * 1019 Jew now this is the difference in energy that results from an electron transition from an n= 5 energy level to an n= 2 energy level so to offer you a short graphic here’s our electron when it relaxes down to the second energy level it’s going to give off a photon of light okay so this is our photon and the energy that corresponds to that photon we have just calculated now given this energy we can rearrange our equation from above where plank’s constant times the speed of light divided by the wavelength of the light is equal to the change in energy and what we’ve done in this step is we’ve rearranged that equation where instead of uh the change in energy being by itself the wavelength is by itself and so we plug in our value for plank’s constant as you can see we’ve wrapped that number in parenthesis we multiply it by the speed of light again wrapped in parenthesis and we divide by our energy our change in energy from above uh now it’s it’s hard to see on the slide here but there is a negative sign in front of this uh 19 that is superscripted and as always we want to make sure that we wrap that number in parenthesis when we do this uh we’re going to get a wavelength of light that is approximately 4.34 * 107 m however we don’t know the visible light range uh along the units of meters so we will immediately convert this to nanometers and we see that we get a wavelength of 434 nanometers and so when we observe the hydrogen discharge tube we’re going to be on the lookout for a wavelength at 434 now let’s take a look at another transition within the Balmer series this is the transition from the n=4 to the n=2 energy level on the previous slide we looked at the transition from the fifth energy level down to the second energy level in this calculation we’re now going to take a look at what the transition will look like between the fourth and the second energy level and as you might expect based on the image here on the screen if I have increasing energy on my y ais then the energy change between the fourth and the second energy levels should be smaller and this is going to have an impact on the observed wavelength of the photon that’s released as the electron relaxes from the excited fourth energy level down to a more relaxed but not yet ground state of the second energy level and again that’s going to give off a photon so again we want to make sure that we label our transition appropriately the fourth energy level is our initial and the second energy level is our final so our final energy calculation will beative k / 2 ^ 2 shown here and we will subtract from that k over 4^2 shown here and if we use the exact values again we get an approximate energy of 4.08 * 1019 jewels uh just a reminder we do not round until we get to the end of our calculation again once we have our change in energy we will then calculate the wavelength of this photon that results from this uh specific change in energy and again we want to make sure that we wrap all of our numbers written in scientific notation in parenthesis and again there is a negative sign here that is just hard to see on the slide so plank’s constant times the speed of light divided by the change in energy gives us an approximate wavelength of 4.86 * 107 m and again our final step is to convert meters to nanometers and this time we see a somewhat larger wavelength at around 486 nanometers for our final calculation we’re going to look at the transition from the third energy level to the second energy level having rehearsed the first two uh calculations with you I now want to encourage you to attempt this calculation on your own first finding the change in energy as a result of moving from the uh final energy level to the initial then we want to use the energy calculated to calculate the wavelength in units of meters and in your final step convert meters to nanometers when we do these three calculations we end up with three different wavelengths and these wavelengths are shown in the image on the screen our first calculated wavelength was a violet wavelength at approximately 434 nanometers our second wavelength was approximately 486 nanometers and this falls within the blue green color range finally in your calculation that you conducted independently you should have come to the answer of about 656 nanometers which is a red line what does this mean the transitions that are occurring within the hydrogen atom are occurring from all of the different energy levels to other energy levels this particular transition series shown on the screen from the fifth the fourth and the third to the second energy level give off a distinct combination of spectral lines that we’re going to attempt to observe there is another transition that we could calculate from the six to the second energy level however that wavelength of light is particularly difficult to observe using our spectroscope so we’re going to uh ignore that line within the Balmer series even though if you were to consult outside sources you may see that there is in fact a fourth line let’s head over to the gas discharge tube and take a look inside of the hydrogen atom

now that we’ve observed the Balmer series within the hydrogen atom I want to remind you of the shortcomings of the bore model of the atom in this part we observe the unique spectral lines that are occurring within a hydrogen atom which only has one electron as the electron is excited it subsequently relaxes releases energy in the form of a photon and we can infer information about the distance between the energy levels by looking at the wavelength of the photon now this does not apply to all other elements because other elements have different numbers of electrons and they have different sized nuclei with different numbers of protons and neutrons and so we will see that we can’t apply the equations used to predict the spectral lines within a hydrogen atom to a helium or a mercury atom but this won’t stop us from observing these discharge tubes and making some very important qualitative and quantitative descriptions about these elements let’s move to helium

here on the screen on the right hand side you can see the helium discharge tube which has a characteristic orange color as you look down through the spectroscope and to the right you will see the spectral lines that result from the helium electron transitions occurring within the subshells of the atom moving left to right we see a few spectral lines in the 450 to the 500 nanometer wavelength range beginning with a light purple and moving towards a blue

and as we continue to move left to right we’re going to see a spectral line approaching 600 pretty sharp line that that just means uh pretty visible 600 nanometer line in yellow and then moving towards the right we can see one very sharp red orange line and one that is a little bit fainter

now the uh spectroscope that we are using uh for this experiment is not going to show a perfect representation of all of the spectral lines within the helium atom but it will do a good job of approximating these wavelengths again each of these distinct wavelengths represents the visible light that is being released when an electron is relaxing from an excited state down to a ground state let’s now take a look at a much larger atom uh the first two hydrogen has only one electron helium having two electrons now we’re going to take a look at the mercury atom whose symbol is Hg and has 80 electrons moving about simultaneously in a beautiful symphony around the nucleus here on the screen on the bottom right hand corner you’re again going to see the uh view of the mercury discharge tube through the spectroscope and it has a a characteristic blue color moving to look to the right you will then observe all of the spectral lines at this point I’m asking you to make qualitative and quantitative descriptions of these lines and to record where you see the approximate spectral lines with respect to our wavelength range beginning with 400 nanometers on the left and 700 nanometers on the right i I ask that you give yourself grace in this task it is not an easy task and approximations are good enough what is unique about mercury is that these spectral lines are the wonderful fingerprint that this element has when the electrons inside of it are excited this gives us a look behind the curtain if you will into the structure of a mercury atom this concludes experiment number two in part one of this experiment we observed the characteristic flame colors of different metal cation solutions in part two of this experiment we then looked behind the curtains into the structure of the atom in the discharged tubes we observed one characteristic color emanating from the lamp and then used the spectroscope to see the individual spectral lines each line indicating the transition of an electron within the atom now as a discussion in part of your lab report I would like for you to answer the following questions question number one relates to part one which metal cation was observed to emit radiation with the shortest and the longest wavelengths which one had the highest frequency and which had the lowest energy to answer these questions you may have to refer back to the equations listed within the beginning of this prelab lecture or the lecture modules itself questions two and three each refer to part two of the experiment while observing the spectral lines for different elements you may have noted that in each gas discharge tube there were different spectral lines what is happening inside of the atom that causes this to emit specific lines in a spectrum and then finally what is the explanation for why hydrogen helium and mercury all had different spectral lines at different wavelengths what does this tell us about the structure of the atom this concludes our time together i hope you’ve enjoyed learning chemistry just as much as I do have a great day