Experiment Seven

Experiment Eight: Atomic Emission: Assigning Transitions for Hydrogen

Prelab Question See the section at the end of the Introduction labeled Pre-Lab Exercise: Graphing, Linear Regression and Least-squares Analysis Using Excel. Follow the instructions for either graphing with PC or MAC. Bring a copy of the graph to hand in.

Purpose Examine and understand the conditions under which hydrogen emission lines fit the Balmer- Rydberg equation. Calculate the emission nf value along with the ionization energy from this nf. Calculate the allowed energy levels of the hydrogen atom.

Introduction

Atomic Emission of Hydrogen and the Balmer-Rydberg Equation An excited electron within a hydrogen atom moves from a low energy (ground state) level to higher energy (excited state) level. This process requires an input of energy (absorption) to promote the electron to the higher energy level. Different wavelengths of light have different energies, thereby promoting a ground state electron to different excited states.

The Rydberg equation describes these electron transitions: 1 1 1  = R  2  2  where ni is the initial electron energy level and nf is the new excited state and  ni n f  R is the Rydberg constant.

When the newly excited electron falls back down to the ground state, light is emitted with an   energy relative to its position from the excited state to the new lower energy level. Not all transitions go back to the ground state, instead stopping somewhere in between at lower energy excited states.

Suppose we have hydrogen emission lines originating from an excited energy level ni and ending in the ground state nf as indicated in Figure 1.

n = 7 n = 6 n = 5

n = 4

n = 3

n = 2

91.2 nm 93.8 nm 95.0 nm 97.3 nm 102.6 nm 121.6 nm

n = 1Figure 1. Ultraviolet Emissions for the Hydrogen Atom

The Rydberg equation as written above describes the promotion of the electron from a lower energy level. That is, the wavelength required to promote an electron into an excited state (the process known as excitation or absorption). The data in Figure 1 is for emission of energy. If the electron is moving down energy levels, then the energy transition is from nf to ni and the energy is emitted as light. The Rydberg equation can be rearranged to a form that describes the emission of light energy 1  R R       2 2    nf ni  According to the new equation, the wavelength of light calculated is emitted light from the electron moving to a lower energy level. With the data in Figure 1. we can graph the wavelengths of the transitions vs ni as a linear equation according to the following statements:

2 Statement #1: A plot of 1/ vs. 1/ni (n = 2, 3, 4…) should be linear with a slope of -R, as indicated by comparing the 2 equations below: 1 R R y = m x + b (equation for a straight line and   2  2  ni nf 1 R y  , m  R and b  Where: 2  nf

 Statement #2: At nf = 1, b = R directly R R m Since at nf = b  2 1, b = R directly and nf =  nf b b Hydrogen Emission Lines in the Visible Region One of the most important experimental observations of the 19th century was that the light given off when substances are vaporized in a flame contains only a few specific colors. Similar emission is observed when an electric current passes through gases. Light from the mercury atoms in a fluorescent lamp, for example, which appears bluish-white to the eye, in fact consists of only four colors: blue, green, yellow and red. The light emitted by an atom (called its spectrum) is characteristic of that atom, no matter what other elements it is combined with. For example, light from sodium vapor is the same as that from sodium chloride, sodium carbonate or sodium hydroxide, while light from hot potassium, or any potassium compound, is distinctly different.

A fundamental question that should be asked about the analysis of atomic emission, is why atoms behave this way. Why do they emit only certain colors of light? The answer lies in the quantization of energy levels. Each atom has electrons of distinct energy, requiring only certain (or quantized) amounts of energy to promote its electrons into higher energy levels. Because of this, only a select number of excited states are available, which equates to only certain emissions of light as electrons fall out of those excited states for lower energy levels within the atom.

Atomic emission, is a useful discovery since it allows unknown atomic substances to be analyzed by observing the light emitted when energized. In fact, the basic act of heating a sample in a flame is enough to view colors. 'Atomic Emission Spectroscopy’ is one of the most sensitive methods available for detecting trace amounts of toxic substances like lead, and the only technique available for identifying the gases found in interstellar space, all based on the light the particles emit specific to their composition and constituent elements.

Spectroscopes In order to see all the colors from an atom’s atomic emission, we need to separate the colors using a prism or diffraction grating. Assembled into a box, we have what is known as a spectroscope. A spectroscope (Figure 2) consists of: (a) a slit - to create a sharp image that is easy to measure; (b) a grating - to separate out the emission colors (very much like the prism); and most importantly (c) a measuring device (ruler) - for identifying the 'lines' at the correct wavelength. Scales could be in m, nm, cm, mm, or Å (angstroms). Figure 2: A spectroscope slit

Grating ruler (look here)

Electromagnetic radiation from the hydrogen discharge lamp enters slit on to the grating. If focused correctly, the grating diffracts (bends) the light according to its wavelength. With the spectroscope off angle, with your eye placed at the grating, the image of the diffracted light is projected onto the ruler. This process is greatly facilitated in a dark room.

Linear Regression and Least Squares Analysis We are going to graph our findings as individual points. The statistical method we are going to use to estimate the quality of our line is called a simple linear regression. Simple linear regression fits a straight line (y = mx + b) through the set of n points in such a way that makes the sum of squared residuals (that is, vertical distances between the points of the data set and the fitted line) as small as possible. See the diagram below. The measure of the “goodness” of our fit is a statistical parameter R (the correlation coefficient)

r3 =

r2 = r1 =

X Figure 1. Least squares residues, r i Figure 3a. Linear regression (best straight line through a series of experimental points), which rarely fit a perfect straight-line. r1, r2 and r3 indicate the residuals. R2 (the correlation coefficient (below) sums the residuals and is an indication of the goodness of the fit.

http://journal.publications.chestnet.org/article.aspx?articleid=1044873

Figure 3b. Linear regression (best straight line through a series of experimental points). The essential parameters in a least squares analysis are the slope (m) = 0.0086, the y-intercept (b) = 0.2459 , and the correlation coefficient (R2) = 0.0806. The correlation coefficient is a measure of the residuals (r). An R2 value of 1.00 indicates a perfect fit (no deviations). Graphing, Standard Curves, Unknowns An important method in any science is the use of graphing to find the relationship (if one exists) between two variable parameters. Graphs are tools we use to aid our understanding of the relationships among variables. Plotting a graph provides us with: • a visual image of the data and their relationships • the ability to predict the results of a change in one variable if the other is held constant In some cases, there is a direct, linear relationship between the parameters. The data obtained in these experiments can be transformed to produce a straight line on a graph described by the equation y = mx + b where m = the slope and b = the y intercept. Making your plot: 1. Which axis is which? The horizontal axis is the independent variable, like time; the vertical axis is the dependent variable, that is, the one that changes in response to the independent variable, like hunger. Hunger increases as a function of time, but time does not increase as a function of hunger. Time would increase regardless of how hungry you feel, so time is the independent, and hunger is the dependent variable. Another way to think about this is that the independent variable is the one you control and vary, plotted on the horizontal axis, and the dependent variable (the one you observe or measure) is on the vertical axis. 2. Label all axes including units, for example: “Time, min.” Use even divisions and whole numbers for your axes. For example, do not use 2.3, 3.1, and 3.6 for your axis divisions; instead, use 2.0, 3.0, and 4.0. 3. Add a title to the graph. Give your plot a good title that indicates the purpose of the graph, or what the graph tells you. Don’t make your title simply the x and y axis labels; a reader can see that already. 4. What scale to use? That is, what numbers should be on each axis? Should the axis begin at 0? How far out should the numbers go? If you are plotting your college GPA as a function of time, the vertical axis does not need to go above 4, nor below zero into negative territory. The horizontal axis, in years, may need to go to 5, or 6, or 7 …… Make sure that you choose a scale that will include all the data points, but don’t choose such a large scale that the curve takes up only a small part of the available graph. Choose scales so that the curve covers the majority of space on the graph. 5. Plot the data points. Then what? There are different procedures on what to do, depending on what it is that you are plotting. If you are plotting your college GPA versus time, then there is no reason to expect that it will be a straight line. We do not expect a linear correlation of the data. In this case, it would be perfectly reasonable to connect the data points. 6. However, if the graph is linear according to theory, draw the best straight line through the points; that means that there will be as many data points above the line as below the line. The line is the best average. For a linear plot y = mx + b, where m is the slope and b is the y- intercept. 7. The slope is the change in y with respect to the change in x: (sometimes called “rise over run”). y-intercept (b) is the value of y when x = 0. 8. Is there a linear correlation in the data on the following plot?

Graph 1. Student Anxiety 5000 Student Anxiety, in SAUs 4000

3000

2000

1000 0 0 1 2 3 4 5 Student time in college, years Pre-Lab Exercise: Graphing, Linear Regression and Least-squares Analysis (Excel MAC) 1. Open an Excel spreadsheet and enter the following in column A and column C. Under the Format Menu (Cells), you will have the option to do superscripts or subscripts as well as write symbols (Symbol font l for lambda) Table 1. Excel Data ni 1/ni2 ( x axis) (nm) 1/y axis) 2 121.5668 3 102.5722 4 97.2537 5 94.9743 6 93.7803

2. Transform the values in column A using a formula. Click in Column B, type “= 1/A2^2” and hit enter. The calculated answer should appear. Copy the formula from Cell B2 to all the other cells in column B. For column D, lets transform using “= 1/C2” and copy to all the other cells in column D. 3. Highlight the cells in column B. Choose format. In the new window, select number and adjust to the decimal to achieve the correct number of significant figures. Use 4 sig fig in Column B and 6 sig fig in Column D. Your data should look like table 2.

Table 2. Excel Data – 1st Transformation

ni 1/ni2 ( x axis) (nm) 1/y axis) 2 0.2500 121.5668 0.008226 3 0.1111 102.5722 0.009749 4 0.0625 97.2537 0.010282 5 0.0400 94.9743 0.010529 6 0.0278 93.7803 0.010663

2 2. Create a plot of 1/ (column D) vs 1/ni (column B) a. Highlight columns B, C and D. b. Click Insert, Chart, XY scatter c. Right click on the white area of the plot and click “select data source”. d. Click on series 1 and series 2 and figure out which one does not belong on your graph then click remove and OK e. Adjust the X and Y-axis by clicking on it, the choose “Make sure each axis has a title, is the correct scale (see figure below) and has the correct number of decimal places. Click “OK’ or hit “Enter.” You can also label these axes and adjust the scale using the toolbar. Click on the chart and open the toolbar. f. Back in the chart, right click on a data point and select “add trendline.” Under type select linear and under options select “Display equation on chart” and “Display R-squared value on chart.” Then OK g. Click on the legend (square on right side) and “delete” it. The graph should open up so it fills the entire box. Move the equation so you can read it clearly. h. Add a title using the toolbar. i. At the end of the above exercise, you should have generated the plot below. j. Print a copy of the chart for your lab notebook. k. With the new chart in the spreadsheet, click in the worksheet to deselect the chart and print two copies of the graph (one to hand in for the pre-lab and one for your lab notebook. Note: If you do not de-select the chart you will print only the chart. You need both the chart (data) and the graph for your lab notebook. Pre-Lab Exercise: Graphing, Linear Regression and Least-squares Analysis (Excel PC) 1. Open an Excel spreadsheet and enter the following in column A and column C. Using the Home tab using the font menu you will have the option to do superscripts or subscripts as well as write symbols (Symbol font l for lambda) Table 1. Excel Data ni 1/ni2 ( x axis) (nm) 1/y axis) 2 121.5668 3 102.5722 4 97.2537 5 94.9743 6 93.7803

3. Highlight the cells in column B. On the home tab choose format then format cells. In the new window under select number select number and adjust to the decimal to achieve the correct number of significant figures. Use 4 sig fig in Column B and 6 sig fig in Column D. Your data should look like table 2. Table 2. Excel Data – 1st Transformation

ni 1/ni2 ( x axis) (nm) 1/y axis) 2 0.2500 121.5668 0.008226 3 0.1111 102.5722 0.009749 4 0.0625 97.2537 0.010282 5 0.0400 94.9743 0.010529 6 0.0278 93.7803 0.010663

2 4. Create a plot of 1/ (column D) vs 1/ni (column B) a. Highlight columns B, C and D. b. Click Insert, Scatter and select the plot that has points only (no lines) c. Right click on the white area of the plot and click “select data”. d. Delete the data series l (nm) by highlighting and clicking on remove and theb click “OK”. e. Adjust the X and Y-axis by clicking on it. Make sure each axis is the correct scale (see figure below) and has the correct number of decimal places. Click “OK’ or hit “Enter.” Label each axis and give the graph a title using the layout tab. f. Back in the chart, right click on a data point and select “add trendline.” Under type select linear and under options select “Display equation on chart” and “Display R-squared value on chart.” Then OK g. Click on the legend (square on right side) and “delete” it. The graph should open up so it fills the entire box. Move the equation so you can read it clearly. h. At the end of the above exercise, you should have generated the plot below. i. print a copy of the chart for your lab notebook. j. With the new chart in the spreadsheet, click in the worksheet to deselect the chart and print two copies of the graph (one to hand in for the pre-lab and one for your lab notebook. Note: If you do not de-select the chart you will print only the chart. You need both the data and the chart for your lab notebook.

Part One: Find the value of R (Rydberg constant) from your prelab data

Procedure 2 1. Your graph is the least squares analysis of 1/ (y-values) vs 1/ni (x-values). The x-axis is 2 -1 1/ni (no units) and the y-axis is 1/ in units of nm . 2. The plot is linear and has a slope (m) = –R in (nm-1).

Data 1. Record your experimentally determined value of R (from your graph) 2. Record your value of the y intercept (from your graph)

Analysis and Calculations 1. Convert your values of R to the SI units of m-1(1/m). 2. Compare your experimental value of R with the literature value of 10,967,758.1 m-1. What statistical parameter do we use to compare two numbers? Do this. 3. From statement #2 (Part A), nf = (R/b) = (-m/b) . Show your calculation of the experimental value of nf. 4. Compare your value of nf to the assumed value of nf = 1. How well does your calculated

value of nf compare with the assumed value of nf = 1? If the two agree, then the data fits the Balmer-Rydberg equation in which the final state is nf =1. If not, then some other nf might apply. What might that value be?

Part Two: Assign and Determine nf Value and the R for Hydrogen Spectrum

Procedure 1. Record the wavelengths observed for the hydrogen emission spectra using the stationary spectroscopes provided for your section. The measurements are within ± 10 nm of the correct value. Note: The violet line is very faint and may be difficult to see. The lines should be distinct. Ask your instructor for help. 2. The textbook values have been included so that you may check the units of your spectroscope. Do not simply copy the values. You will not get exactly the same readings, but they should be close. The following table should be in your lab notebook:

Color Wavelengths ( Lit. values (nm) violet 410.1 violet - blue 434.1 blue - green 486.1 red 656.3

Analysis and Calculations If the data agrees with the theory (Balmer-Rydberg formula), then a plot of 1/ vs. 1/ni2 should be linear with a slope of -R . From Part One, we assumed transitions based on Figure 1, where nf = 1. The only way to know if our transitions have an nf of 1 is to plot the data for different values of nf and see which one gives the 'best fit' based on linear regression of the data. 1. Copy the following four tables in your lab notebook. 2. Fill in tables 4 – 6 using the wavelengths measured with the spectroscopes. 3. Using the assigned ni values and your wavelengths, plot the data as in the prelab exercise. 4. Analyze the data by linear regression and use the equation provided from the linear regression from each set of data to complete Table 7 in your lab notebook. Record the slope (m) , y- intercept (b) and least square value (R2) in the table. 5. Based on the fact that the quantum number nf must be an integer, calculate the value of nf afforded by each set of data. ONLY ONE SET OF DATA WILL PROVIDE AN INTEGER. The formula below will be useful. R R m b  2  nf =  n f b b 5. Include copies of your graphs in your lab report .

Table 4. Set #1; nf = 1; ni = 2, 3, 4, 5   ni values 1  1/ 2 ni ni = 2 (red) ni = 3 (blue-green) ni = 4 (violet-blue) ni = 5 (violet)

Table 5. Set #2; nf = 2; ni = 3, 4, 5, 6 ni values 1  1/ 2 ni ni = 3 (red) ni = 4 (blue-green) ni = 5 (violet-blue) ni = 6 (violet)

Table 6. Set #3; nf = 3; ni = 4, 5, 6, 7 ni values 1  1/ 2 ni ni = 4 (red) ni = 5 (blue-green) ni = 6 (violet-blue) ni = 7 (violet)

Table 7. Least-Squares Data 2 Set m b R value nf Set #1 Set #2 Set #3 Post-Lab Questions 1. Which set of data provided a whole number for nf? 2. Was the liner-regression analysis useful in this determination? Do not just answer yes or not. Explain why it was useful or why not. Use complete sentences. Explain yourself. 3. What was the value of R, the Rydberg constant, for set #2? 4. Using your book and the internet, briefly explain what differences would be seen comparing the atomic emission of hydrogen to that of neon. What new colors are visible? What transitions are occurring for these new emissions?