Forensic Spectral Anaylysis: Warm up!
The study of triangles has been done since
ancient times. Many of the early discoveries
about triangles are still used today. We will only
be concerned with the "right triangle" (see
below) which is a triangle with one of the three
angles being 90 degrees. We can use these
triangles to determine the heights of buildings,
survey property boundaries and determine the
length of the waves in the spectrum of colors.
Even the heights of mountains on the moon can
be determined.
This exercise will apply the Pythagorean
theorem to calculating wavelengths of light that
you will measure. You also will need to do a
percent calculation to estimate the error in your
result as illustrated below.
This illustration above is the general right triangle whose sides have lengths called A, B and C . The angle A is the value in degrees
of the angle illustrated.
Pythagorean theorem: The ancient Greek mathematical schools discovered that the hypotenuse (side "C" above) of the triangle as
well as the sides of the triangle ("A" and "B") are all wonderfully connected in a simple formula.
The hypotenuse squared is equal to the sum of the square of each side. This is known as the Pythagorean theorem and can be represented
by the following formula:
C2 = A2 + B2
ILLUSTRATIVE EXAMPLE: Given: A = 3 and B = 4 if we now want the value of C=?
C2 = 32 + 42 = 9 + 16 = 25 or
C2 = 25 To find the value of "C" we take the square root of the number since C "squared" means what number times itself will give 25
1/2
in this case. The answer in this case is 5. That is, 5 x 5 = 25 or squareroot 25=(25) =5 or we would state that C = (25) 1/2 = 5
is the answer for the hypotenuse of a triangle whose sides are 3 and 4. or 5 x 5 =25!
ILLUSTRATIVE EXAMPLE: USING THE WINDOWS CALCULATOR set in “View” as Scientific FOR THIS EXAMPLE
WE COULD PROCEED AS FOLLOWS (you can use your own calculator)
Enter the 3 and click X^2 key then store in memory with the MS key
Enter the 4 and click X^2 key then add to memory with the M+ key
Recall the Memory with the MR key (you should have the number 25)
Then take the square root by clicking the INV (inverse operation) and the X^2 key. If all goes well you will get a 5 for the answer
other portable calculators have M+ and MR and X 2 keys but the square root function has it’s own key
QUESTION 12. USE A SCIENTIFIC CALCULATOR FOR THE FOLLOWING
find C?
C=?
Given a triangle whose sides are A = 8 and B = 13
The Wavelength( in very small units called Nanometers or nm) of one of the colors produced by a Helium emission spectrum
using a device that sets up a right triangle of physical variables is given ultimately as Wavelength =1370.4 x B/C in nm!
ILLUSTRATIVE EXAMPLE:Wavelength =1370.4 x B/C nm with C = ( A2 + B2 )1/2 by the Pythagorean
theorem
2
2
: given A=20 B=10 then C= ( 20
+ 10
)1/2 = (400+100) 1/2 = (500)1/2 =22.36
Hence Wavelength =1429.6 x B/C= 1370.4 x 10 /22.36 = 1370.4 x 0.4472 =612.88 nm units of wavelength.
QUESTION14.: Given A =20 B=12
Wavelength =?
Solve for THE Wavelength using the formula above:
Problem
How can we measure the wavelengths of radiation emitted by elements and use them to identify the element?
We will explore the colors produced by elements being excited by high voltages (DANGER :DO NOT HANDLE OR TOUCH THE
SPECTRUM TUBE APPARATUS while you are touching any plumbing or metal grounds BECAUSE OF THE SHOCK HAZARD.
The INSTRUCTOR will demonstrate the correct SAFE handling of the equipment. The dispersion of the colors is produced by
a special transparent film called a diffraction grating which looks like a 35 mm glass slide but behaves like a prism or the raindrops
that create the rainbow.. See the picture below.
Apparatus: Construct the meter stick spectroscope as pictured below: Use the instructors model and this image as a
guide. The spectroscope consists of the modified meter stick optical bench: includes meter stick, supports, a diffraction grating,
holder with grating(screen holder), a graduated cross stick and holder. Note: Cross stick has an opening slit in the center!
Needed Information and Skills
It was discovered last century that every element and many compounds will, upon being excited to glow by flame or electricity
partial vacuum, produce a unique pattern of colors. Thus, we have a "fingerprint" of an element which can be used to find out
whether or not an element is present. How much of the element is present is indicated by the brightness of the unique colors. The
brighter the color the greater the amount of the element present. This technique helps scientists determine the composition of
many objects. The principles behind the technique you use here is used in medicine to learn about the composition of your blood
and, hence,the state of your health. It is also used in criminology to examine various aspects of evidence in crimes and, hence,
help solve the crime. The Astronomer analyzes the colors of stars and is able to determine the chemical composition of the Stars
as well as motions, temperature etc. and ultimately determine the evolution and fate of stars like our sun.
We will be using spectra of colors from excited gases that are called the Emission Spectra . The laboratory contains a
SPECTRUM CHART that shows emission spectra from various gases and other types of spectra you encountered in the
preparation assignment. The theory on how the grating spreads out light into a spectra depends on the fact that each grating has
in our case 600 lines /mm engraved into them. Each of these lines acts as a source of the original light source resulting in 600
images being projected together. This results in waves constructively and destructively interfering with each other. The result is the
separation of the colors making up the original source both to the left and right of the original source. You might think of the grating
as a device like the prism that separates (disperses) the colors of light produced by glowing objects.
We note that the colors within light are waves and are assigned wavelengths (distance between crests of wave). These
lengths are very small and are measured as one BILLIONTH of a METER or 10-9 METERS. We give this small unit a name called
a NANOMETER ABREVIATED “nm”. hence, 1 nm = 10-9 meter. Refering to your charts various red colors are around 650.0
NANOMETERS and Violets are around 400 nm in the size of their wavelengths. NOTE: The laboratory charts are in a 10-10 meter
unit known as an angstrom. Hence, 1nm = 0.1 angstrom or in other words the laboratory charts express visible light as 4000 to
7000 angstrom which in nanometers is 400 to 700. Hence, Just divide any 4 digit angstrom unit by 10 to get a three digit number.
USING THE METER STICK SPECTROSCOPE:
One look through the grating and through the hole in the cross-stick and centers the gas tube in the opening.
The instructor will demonstrate the technique that is illustrated in the following schematic figure.
Study this figure keeping in mind the following points:
We note in this schematic your eye (see figure) lines up the Spectrum tube image with the OPENING IN THE center of cross
stick. Hence, you will use the open slit for aiming. The spectrum of colors are spread out along the cross-stick on the left and right
side of the emission tube you aim at. Further note that this exercise is set to work for a distance of 1 meter from the tube to the
meter stick as shown in the figure. The colors are the same on either side of the tube image except that the colors closer to the
tube are the violet end of the spectrum and the ones further away are the red end. WE WORK WITH THE SPECTRUM TO THE
RIGHT (see figure) OF THE TUBE RUNNING FROM VIOLET TO RED IN THIS EXERCISE.
In summary: the diffraction grating will split the source of light into two parts. When you look through the grating on the optical
bench will see image of the original light source with sets of color images of the source on either side. It is important to center on
the meter stick the image of the original spectrum tube when making measurements. The amount of separation seen along the
cross-stick called “B” distance in the figure above of the individual color from the main slit image can by theory beyond this course
give us a value for the wavelength. The accurate measuring of the wavelength values determines what element is present in our
sample under observation.
CALCULATION of WAVELENGTH:
The formula for wavelength in NANOMETER units for our diffraction grating apparatus is given from the optical theory of
interference in Physics as
Wavelength =1429.6 x B/C
In units of nm
with C = √ A2 + B2 by the Pythagorean theorem as you can see from the figure We always keep A = 20 centimeters and
measure B with our observations. Then we can use this formula to get the wavelength which we do later for the case of
helium below.
PROCEDURE TO MEASURE THE WAVELENGTHS OF HELIUM
10.Check the setup of the meter stick spectroscope . It is important to set the distance between the cross stick and the grating to
20.0 cm. Thus. the distance ’A’ in the figure is fixed at 20 cm or A=20.0 for what follows.
BE CAREFUL NOT TO HURT YOUR NEIGHBOR WITH THE METER STICKS BY ACCIDENTALLY POKING THEM.
I: “B” VALUES OF THE TABLE BELOW
Reviewing the spectroscope: The grating is in the screen holder. The cross- stick is at right angles to the meter stick optical
bench. A = 20 centimeters. USE THE CENTER SLIT OPENING ON THE CROSS-STICK TO AIM THE METER STICK
DIRECTLY AT THE CENTRAL IMAGE OF THE TUBE.. The tube source should be about one meter from the cross-stick side of
the optical bench. Center the image of the tube and look for first emission line to the right. .MAKE SURE OTHER LIGHT
SOURCES ARE NOT IN YOUR LINE OF SIGHT OTHERWISE YOU WILL GET LINES FROM THEM!!!!! It helps if you have a
partner jot down the values of B for all colors. Which you can place in the table below.
NOTE: Take the B MEASUREMENTS FOR THE FOLLOWING COLORS VIOLET, BLUE, STRONG GREEN, YELLOW, STRONG
RED, WEAK RED. IGNORE THE WEAK GREEN LINE FOR IT IS NOT ON YOUR CHARTS.
For example. If you match A COLOR to the 9 cm. line the distance B is 9 cm. Record this B value in the data table below. You
might also note all the other B values for each color at this point. These will range from 7 to 15 cm.
II: C VALUES Compute the value of C(see figure) by the Pythagorean theorem C = √ A2 + B2
data table under C column.
and record the value(s) in the
III: COMPUTE THE WAVELENGTH OF EACH COLOR: the Wavelength IN NANOMETERS by Wavelength =1370.4 x B/C
Record your computed value in the WAVELENGTH column of the data table. This is your computed value.
IV; GET THE KNOWN WAVELENGTH VALUES FROM THE CHART: To see how good it is we do the next two steps and get a
percent error from the chart(TRUE) value. Consult the large spectrum chart and carefully estimate the value of the wavelength for
the yellow line and other colors of Helium and record the values in the CHART:WL column. Thus, CHART:WL is the variable
containing the value you read off the chart!
V COMPUTE THE % ERROR OF THE WAVELENGTHS: by the following formula
WAVELENGTH - CHART:WL
---------------------------------------- X 100
CHART:WL
% ERROR =
HELIUM DATA TABLE
1'
2'
COLOR
YUCH (EX)
A=20 CM
3'
B
9.2
C = √ A 2 + B2
√484.64=21.01
computed
4'
B/C
.4379
5'
Wavelength
=1370.4 x B/C
600.1
note: negative means Chart:WL> Wavelength
chart
6'
WAVELENGTH
from CHART
629.5
7'
% ERROR
=(5' -6')/6' X 100
-4.6% ("-" means
your value < chart)
NOTE THE 6 COLORS ARE: VIOLET, BLUE(DIM), GREEN, YELLOW, RED, RED(DIM)
Forensic Spectral Analysis: Draw the main lines of the spectrum of the unknowns below.
USE THE CHART AND LINK BELOW TO IDENTIFY THE ELEMENT (Link also in class outline)
Calculate two primary lines and use the charts to identify the unknown elements or molecules
NOTE SPECTRUM WITH THE VIOLET SIDE TO THE LEFT AND RED SIDE TO THE RIGHT! Mark colors and colored
lines in appropriate color areas in the rectangular boxes below.
Unknown 1
Wavelengths of 2 primary lines =
400 nm Violets
Blues
ID=_______________________
Greens
Yellows
Oranges
Reds
Unknown 2 Wavelengths 2 primary lines =
400 nm Violets
Blues
ID=_____________________
Greens
Yellows
Oranges
Reds
Unknown 3 Wavelength of 2 primary lines=
400 nm Violets
Blues
750 nm
750 nm
ID=___________________
Greens
Yellows
Oranges
Reds
750 nm
)
Unknown 4 Wavelengths of 2 primary lines=
400 nm Violets
Blues
Greens
ID=________________
Yellows
Oranges
Reds
750 nm
|
Unknown 5 Wavelengths of 2 primary lines =
400 nm Violets
Blues
Greens
ID=________________
Yellows
Oranges
Unknown 6 Wavelengths of 2 primary lines=
400 nm Violets
Blues
Greens
Reds
750 nm
ID=_________________
Yellows
Oranges
LINK to ATOMIC SPECTRA ON THE WEB. Click on a spectra for it's details!
Reds
750 nm
http://astro.u-strasbg.fr/~koppen/discharge/
Afterthought questions Do these questions after you do the lab.
Use the index in "Kutner" and Google if needed to find answers
Question 1: Considering Light as a wave the distance between one crest of the wave to the next
is called the ?
Question 2: Approcimately what color would you be observing if you measured a wavelength
of light at 600 nm (nano-meters).
Question 3: If you measured a wavelength of 10 m (METERS) then you would be observing
what type of radiation?
Question 4: as in question 3 a wavelength of 10-10 METERS would be what type of radiation?
Question 5: Name two other type of radiations besides light and the types in questions 3 and 4
that are part of the spectrum (electromagnetic radiation) ?
Question 6 : The band of color from atoms emitting photons of all colors is called a ?
Question 7: True or False? Each element can emit only certain wavelengths
Question 8. What part of the atom falls down to a lower energy level or ground state also
emitting a photon of a specific wavelength (color).?
We say it makes a transition from one orbit to another
Question 9: Define an absorption spectrum
Question 10: The wavelength of the Hydrogen alpha (RED) emission line is ? nm?.
Question 11: The wavelength of the Hydrogen line in the emission spectrum that is Blue in
color is called Hydrogen Gamma and has a wavelength of ? nm.