PHYSICS LESSON 19 FOR
THURSDAY, JULY 01, 2010
I Introduction
II Logistics: Return of Papers, Running Grades, Turn in Assignments Due, etc. and
III Test 5
IV Mind Game 5
V Homework – no more
VII Essay 5: Enrico Fermi, due 7/01
End of course
Friday, July 2, 2010
Wednesday, June 30, 2010
SOLUTION SET 5
1. A wave has crests and troughs. From the top of one crest to the bottom of the next trough is 13 cm. The length from crest to trough along the x axis is 28 cm. Find the a. amplitude and b. wavelength.
Solution:
a. Since the top of one crest to the bottom of the next trough is 13 cm, and since the distance from the center (zero) line to the apex (crest, top, etc.) is the Amplitude, and since the center line is halfway from the apex to the trough, we multiply the 13 cm by ½, or, ½ (13 cm) = A = 6.5 cm.
b. Since the length from crest to trough along the x-axis is 28 cm, and since the definition of wavelength the distance from crest to crest, and the distance from crest to trough is only half a wavelength, then we have to double the distance given to find the wavelength, l = 2 x 28 = 56 cm.
2. The speed of surface waves in water decreases as the water becomes more shallow. Imagine water waves crossing the surface of an otherwise calm lake at a speed of v1 = 2.0 m/s, with a wavelength of l1 = 1.5 meters. When these same waves approach shore, the speed decreases to v2 = 1.6 m/s, while the frequency remains the same. Find the wavelength, l2, in the shallow waters.
Solution: We have learned that by multiplying the wavelength, l, by the frequency, n, we get that “famous” equation, resulting in the velocity: l n = v. To find the non-changing frequency, we can rewrite the equation as n = v1/l1.= (2.0)/(1.5) = 1.33 Hz. Remember, n 1 = n 2 = n = 1.33 Hz. So, using the same equation that l n = v., to find, , we just divide both sides by thfrquency, n. And the get this: l 2 = v2 /n.= (1.6 m/s)/(1.33) = 1.2 meters.
3. Suppose that you wanted to double the wave speed, from v to 2v, on a tight string. The force that is keeping it tight is called “tension” but it is still a force and has units of Newtons. How much would you have to increase the tension so the velocity would double?
Solution: We remember from the laboratory exercise, “Tin Can Phone with Soundwaves on a String,” that the velocity is: v = (T/m)½ , where T is the tension or force, and m is the linear density. So, if we wanted the velocity to go from v to 2v, then we would have to quadruple the tension (i.e., four times more, because: 2V = (4T/d)½
4. Waves on a particular string travel at v1 = 16 m/s. To double that speed to v2 = 32 m/s, how much would you have to increase the tension?
Solution:
The clever student should quickly realize that this is the same identical problem as Number 9, so the answer is, again, . quadruple the tension.
5. Write an expression for a harmonic wave, such that amplitude, A = 0.16 m, wavelength l = 2.1 m, and period of P = 1.8 seconds.
Solution: On page in the book, it says that the expression for a harmonic wave is:
y = A cos(2px/l – 2pt/P), and thus, if you plug in the data given, your answer becomes:
y(x,t) = 0.16 Cos[(2px/2.1) – (2pt/1.8)]
6. A soundwave in air has a frequency of n1 = 425 Hz.
a. Find its wavelength
b. If its frequency is increased, does the wavelength increase, decrease, or remain unchanged
c. Calculate its wavelength if the frequency is n2 = 475 Hz.
Solution: Once again, we use the “famous” equation, l n = v., and to find wavelength, we divide both sides by the frequency. To get l = v/n.
a. l = v/n. = (342 m/s)/(425 Hz) = 0.805 meters. (but where did we get the speed, v? This is a sound wave. In air. The speed of sound in air is always 342 m/s.
b.l n = vsound
c. l = 342/475 = 0.72 meters.
7. A man throws a rock down to the bottom of a well. From the instant that it leaves his hand until the moment he hears the rock hit bottom, a period of 1.20 seconds passes. The depth of the well is 8.85 meters. Find the initial speed, vi, of the rock (the instant that it leaves the man's hand).
Solution: We are looking for vi, the initial speed of the rock (the instant that it leaves the man's throwing hand). So, what do we know? We know that the final speed of the rock, vf = 0.0 m/s (it is stopped when it hits the well's bottom). Maybe we can use the formula:
vf2 - vi2 = 2 a y, where since vf = 0.0, we can re-write it as:
- vi2 = 2 a y. We know the number “2.” We know the distance down, y = 8.85 meters, but do we know the acceleration, a?
What is “a” ? a = g + a’. And g is Earth's gravitational pull, while a' is the extra acceleration added by the man's hand. Since I don't know where this is going, I am going to put this path to solving it on “hold” and try something else.
We know from a previous lab that the average speed as it falls is related to the initial speed, i.e., vave = ½ vi
We need to find the average velocity. We know that speed is distance over time, or, v = y/t. Let's call:
t1 = the time it takes for the object to reach the bottom after being thrown
t2 = the time for the object’s sound to reach the top after making noise at the bottom.
We do know that total time from the time the rock was released until the sound reached the man's ears is given: t1 + t2 = 1.2 s.
We know that the speed of sound here is: vs = y/t2 ; so, t2 = y/vs = 8.85 m/342 m/s = 0.026 s. If t2 = 0.26 seconds, then t1 can be found by subtraction.
t1 + 0.026 = 1.2 s
t1 = 1.2 s - 0.026 = 1.174 s
Now we can find the average speed, as the rock was falling/thrown down: distance/time, or y/t1, or vave = 8.85/1.174 = 7.54 m/s
and, if vave = ½ vi, then 2vave = vi..= (2)(7.54) = 15.08 m/s. Rounding, we get:
vi = 15 m/s.
8. Twenty identical violins are being played by 20 identical violinists at the same identical volume, for an aggregate sound level of 82.5 dB.
a. Find the dB level of one of the violins.
b. Find the dB level of twice as many identical violins (40).
Solution: According to the table provided in the book (or on the test), one sees that a dB level of 80 has an energy flux of 10-4 Watts/m2.
a. If 20 violins has an energy flux of 10-4 Watts/m2, then one violin would have 1/20th of that, or, 10-4 /20 = 100 x 10-6 / 20 = 5 x 10-6. Rechecking the table shows that an energy flux of that amount would be the equivalent to about 66 dB.
b. If 20 violins has an energy flux of 10-4 Watts/m2, then 40 violins would have twice that, or, 2 x 10-4 Watts/m2, and checking the chart, that energy flux would give that a reading of about 85 dB.
9. Find the weight (in Newtons) of the air in your physics class. Assume that the room is a cube of 4 meters on a side, and that the density of air is rair = 1.29 kg/m3.
Solution:
Volume 4 x 4 x 4 = 64 m3 ; rair = 1.29 kg/m 3. The density, r, is defined to be:
r = mass/volume.
So, the mass = r vol = (1.29)(64) = 82.56 kg. Finally, the “weight” in Newtons is its force in gravity: F = m g = (82.56)(9.8) = 809 Newtons.
10. You buy a gold ring at a pawn shop, and run a “test” to determine if it's pure gold. The ring's mass is m = 0.014 grams, and has a volume of 0.022 cm3. The density of gold is rAu = 19.3 kg/m3.
Solution: Density is mass divided by volume, and it this case, r = M/V = (0.014g/0.0022 cm3). = 6.6 g/cm3. However, since rAu = 19.3 kg/m3. You can be sure that the ring is NOT solid gold.
11. Crutches will often have rubber tips at the bottom, so that contact with a flooring surface will decrease scratching of the floor, increase friction, and its broader tip spreads the pressure out over a larger area. Assume the radius of the tip of the cane to be 1.2 cm, and that the radius of the rubber tip is 2.5 cm. What is the ratio of the pressure of the cane alone, P1, to the pressure with the rubber tip, P2: P1/P2.
Solution:
Pressure = Force/Area
Area p r2
Area 1 = p r2
Area 2 = p (2.5/1.2)2r2 =p (2.08)2r2
Thus, Area 2 is 4.34 greater than Area 1, and thus, the pressure in Area 2 is 1/(4.34) as great; in other words, is 0.23 the pressure. P1/P2. = 4.34.
12. Two drinking glasses, G1 and G2, are filled with water to the same level of 5.0 cm. The diameter of the base of G1 is twice the diameter of the base of G2, but otherwise the two glasses are identical.
a. Is the weight of the water in G1 greater, less, or the same as weight in G2?
b. Is the water pressure at the bottom of G1 greater, less, or the same as the water pressure at the bottom of G2?
Solution: There is more water in G1 than G2, but the same area
a. G1 has a greater weight.
b. Water pressure is identical.
13. On Wednesday, August 15, 1934, William Beebe and Otis Barton made history by descending in the Bathysphere, a steel sphere 4.75 feet in diameter, to 3028 feet below sea level.
a. as the Bathysphere was lowered, was the buoyant force exerted on it at a depth of 10 feet greater, less, or equal to it at 50 feet?
b. Choose the best explanation from below.
I Magic
II Miracle
III Mayonnaise.
Solution:
a. Same
b. III
14. A fish called Wanda is carrying a pebble in its mouth swims with a small constant velocity in a small bowl. When the fish drops the pebble to the bottom of the bowl, does the water level rise, fall or stay the same?
Solution: Stays the same.
15. Explain what a a hydrometer is.
Solution:
How would I know? I'm not an auto mechanic.
A hydrometer is an instrument used to measure the relative density of liquids to water, which has a density of r = 103 kg/m3. It comes from two words, hydro, meaning “water,” and metros meaning “to measure.”
Hydrometers are typically made of glass, which, itself is a liquid. The device consists of a cylindrical stem and a bulb weighted with the element mercury (not the planet or the car or the Roman god). This allows the stem and bulb to float upright. The liquid to be tested (such as honey or antifreeze) is poured into a tall jar, and the hydrometer is gently lowered into the liquid – the honey or antifreeze, etc., until it floats freely. The point at which the surface of the liquid touches the stem of the hydrometer is noted. Hydrometers usually contain a paper scale inside the stem, so that the relative density can be read directly. The scales may be of different types, as hydrometer units have not yet been standardized by the SIU.
Hydrometers may be calibrated for different uses, such as a lactometer for measuring the density (creaminess) of milk, a saccharometer for measuring the density of sugar in a liquid, or an alcoholometer for measuring higher levels of alcohol in liquor and other distilled spirits.
END
Solution:
a. Since the top of one crest to the bottom of the next trough is 13 cm, and since the distance from the center (zero) line to the apex (crest, top, etc.) is the Amplitude, and since the center line is halfway from the apex to the trough, we multiply the 13 cm by ½, or, ½ (13 cm) = A = 6.5 cm.
b. Since the length from crest to trough along the x-axis is 28 cm, and since the definition of wavelength the distance from crest to crest, and the distance from crest to trough is only half a wavelength, then we have to double the distance given to find the wavelength, l = 2 x 28 = 56 cm.
2. The speed of surface waves in water decreases as the water becomes more shallow. Imagine water waves crossing the surface of an otherwise calm lake at a speed of v1 = 2.0 m/s, with a wavelength of l1 = 1.5 meters. When these same waves approach shore, the speed decreases to v2 = 1.6 m/s, while the frequency remains the same. Find the wavelength, l2, in the shallow waters.
Solution: We have learned that by multiplying the wavelength, l, by the frequency, n, we get that “famous” equation, resulting in the velocity: l n = v. To find the non-changing frequency, we can rewrite the equation as n = v1/l1.= (2.0)/(1.5) = 1.33 Hz. Remember, n 1 = n 2 = n = 1.33 Hz. So, using the same equation that l n = v., to find, , we just divide both sides by thfrquency, n. And the get this: l 2 = v2 /n.= (1.6 m/s)/(1.33) = 1.2 meters.
3. Suppose that you wanted to double the wave speed, from v to 2v, on a tight string. The force that is keeping it tight is called “tension” but it is still a force and has units of Newtons. How much would you have to increase the tension so the velocity would double?
Solution: We remember from the laboratory exercise, “Tin Can Phone with Soundwaves on a String,” that the velocity is: v = (T/m)½ , where T is the tension or force, and m is the linear density. So, if we wanted the velocity to go from v to 2v, then we would have to quadruple the tension (i.e., four times more, because: 2V = (4T/d)½
4. Waves on a particular string travel at v1 = 16 m/s. To double that speed to v2 = 32 m/s, how much would you have to increase the tension?
Solution:
The clever student should quickly realize that this is the same identical problem as Number 9, so the answer is, again, . quadruple the tension.
5. Write an expression for a harmonic wave, such that amplitude, A = 0.16 m, wavelength l = 2.1 m, and period of P = 1.8 seconds.
Solution: On page in the book, it says that the expression for a harmonic wave is:
y = A cos(2px/l – 2pt/P), and thus, if you plug in the data given, your answer becomes:
y(x,t) = 0.16 Cos[(2px/2.1) – (2pt/1.8)]
6. A soundwave in air has a frequency of n1 = 425 Hz.
a. Find its wavelength
b. If its frequency is increased, does the wavelength increase, decrease, or remain unchanged
c. Calculate its wavelength if the frequency is n2 = 475 Hz.
Solution: Once again, we use the “famous” equation, l n = v., and to find wavelength, we divide both sides by the frequency. To get l = v/n.
a. l = v/n. = (342 m/s)/(425 Hz) = 0.805 meters. (but where did we get the speed, v? This is a sound wave. In air. The speed of sound in air is always 342 m/s.
b.l n = vsound
c. l = 342/475 = 0.72 meters.
7. A man throws a rock down to the bottom of a well. From the instant that it leaves his hand until the moment he hears the rock hit bottom, a period of 1.20 seconds passes. The depth of the well is 8.85 meters. Find the initial speed, vi, of the rock (the instant that it leaves the man's hand).
Solution: We are looking for vi, the initial speed of the rock (the instant that it leaves the man's throwing hand). So, what do we know? We know that the final speed of the rock, vf = 0.0 m/s (it is stopped when it hits the well's bottom). Maybe we can use the formula:
vf2 - vi2 = 2 a y, where since vf = 0.0, we can re-write it as:
- vi2 = 2 a y. We know the number “2.” We know the distance down, y = 8.85 meters, but do we know the acceleration, a?
What is “a” ? a = g + a’. And g is Earth's gravitational pull, while a' is the extra acceleration added by the man's hand. Since I don't know where this is going, I am going to put this path to solving it on “hold” and try something else.
We know from a previous lab that the average speed as it falls is related to the initial speed, i.e., vave = ½ vi
We need to find the average velocity. We know that speed is distance over time, or, v = y/t. Let's call:
t1 = the time it takes for the object to reach the bottom after being thrown
t2 = the time for the object’s sound to reach the top after making noise at the bottom.
We do know that total time from the time the rock was released until the sound reached the man's ears is given: t1 + t2 = 1.2 s.
We know that the speed of sound here is: vs = y/t2 ; so, t2 = y/vs = 8.85 m/342 m/s = 0.026 s. If t2 = 0.26 seconds, then t1 can be found by subtraction.
t1 + 0.026 = 1.2 s
t1 = 1.2 s - 0.026 = 1.174 s
Now we can find the average speed, as the rock was falling/thrown down: distance/time, or y/t1, or vave = 8.85/1.174 = 7.54 m/s
and, if vave = ½ vi, then 2vave = vi..= (2)(7.54) = 15.08 m/s. Rounding, we get:
vi = 15 m/s.
8. Twenty identical violins are being played by 20 identical violinists at the same identical volume, for an aggregate sound level of 82.5 dB.
a. Find the dB level of one of the violins.
b. Find the dB level of twice as many identical violins (40).
Solution: According to the table provided in the book (or on the test), one sees that a dB level of 80 has an energy flux of 10-4 Watts/m2.
a. If 20 violins has an energy flux of 10-4 Watts/m2, then one violin would have 1/20th of that, or, 10-4 /20 = 100 x 10-6 / 20 = 5 x 10-6. Rechecking the table shows that an energy flux of that amount would be the equivalent to about 66 dB.
b. If 20 violins has an energy flux of 10-4 Watts/m2, then 40 violins would have twice that, or, 2 x 10-4 Watts/m2, and checking the chart, that energy flux would give that a reading of about 85 dB.
9. Find the weight (in Newtons) of the air in your physics class. Assume that the room is a cube of 4 meters on a side, and that the density of air is rair = 1.29 kg/m3.
Solution:
Volume 4 x 4 x 4 = 64 m3 ; rair = 1.29 kg/m 3. The density, r, is defined to be:
r = mass/volume.
So, the mass = r vol = (1.29)(64) = 82.56 kg. Finally, the “weight” in Newtons is its force in gravity: F = m g = (82.56)(9.8) = 809 Newtons.
10. You buy a gold ring at a pawn shop, and run a “test” to determine if it's pure gold. The ring's mass is m = 0.014 grams, and has a volume of 0.022 cm3. The density of gold is rAu = 19.3 kg/m3.
Solution: Density is mass divided by volume, and it this case, r = M/V = (0.014g/0.0022 cm3). = 6.6 g/cm3. However, since rAu = 19.3 kg/m3. You can be sure that the ring is NOT solid gold.
11. Crutches will often have rubber tips at the bottom, so that contact with a flooring surface will decrease scratching of the floor, increase friction, and its broader tip spreads the pressure out over a larger area. Assume the radius of the tip of the cane to be 1.2 cm, and that the radius of the rubber tip is 2.5 cm. What is the ratio of the pressure of the cane alone, P1, to the pressure with the rubber tip, P2: P1/P2.
Solution:
Pressure = Force/Area
Area p r2
Area 1 = p r2
Area 2 = p (2.5/1.2)2r2 =p (2.08)2r2
Thus, Area 2 is 4.34 greater than Area 1, and thus, the pressure in Area 2 is 1/(4.34) as great; in other words, is 0.23 the pressure. P1/P2. = 4.34.
12. Two drinking glasses, G1 and G2, are filled with water to the same level of 5.0 cm. The diameter of the base of G1 is twice the diameter of the base of G2, but otherwise the two glasses are identical.
a. Is the weight of the water in G1 greater, less, or the same as weight in G2?
b. Is the water pressure at the bottom of G1 greater, less, or the same as the water pressure at the bottom of G2?
Solution: There is more water in G1 than G2, but the same area
a. G1 has a greater weight.
b. Water pressure is identical.
13. On Wednesday, August 15, 1934, William Beebe and Otis Barton made history by descending in the Bathysphere, a steel sphere 4.75 feet in diameter, to 3028 feet below sea level.
a. as the Bathysphere was lowered, was the buoyant force exerted on it at a depth of 10 feet greater, less, or equal to it at 50 feet?
b. Choose the best explanation from below.
I Magic
II Miracle
III Mayonnaise.
Solution:
a. Same
b. III
14. A fish called Wanda is carrying a pebble in its mouth swims with a small constant velocity in a small bowl. When the fish drops the pebble to the bottom of the bowl, does the water level rise, fall or stay the same?
Solution: Stays the same.
15. Explain what a a hydrometer is.
Solution:
How would I know? I'm not an auto mechanic.
A hydrometer is an instrument used to measure the relative density of liquids to water, which has a density of r = 103 kg/m3. It comes from two words, hydro, meaning “water,” and metros meaning “to measure.”
Hydrometers are typically made of glass, which, itself is a liquid. The device consists of a cylindrical stem and a bulb weighted with the element mercury (not the planet or the car or the Roman god). This allows the stem and bulb to float upright. The liquid to be tested (such as honey or antifreeze) is poured into a tall jar, and the hydrometer is gently lowered into the liquid – the honey or antifreeze, etc., until it floats freely. The point at which the surface of the liquid touches the stem of the hydrometer is noted. Hydrometers usually contain a paper scale inside the stem, so that the relative density can be read directly. The scales may be of different types, as hydrometer units have not yet been standardized by the SIU.
Hydrometers may be calibrated for different uses, such as a lactometer for measuring the density (creaminess) of milk, a saccharometer for measuring the density of sugar in a liquid, or an alcoholometer for measuring higher levels of alcohol in liquor and other distilled spirits.
END
LESSON 18
PHYSICS LESSON 18 FOR
WEDNESDAY, JUNE 30, 2010
I Introduction
II Logistics:
a. Return of papers
b. Running grades
III Review Lesson 17: Spectroscopy
Isaac Newton
Gas hydrogen
_____ n = 2
__e__ n = 1 ground state give off light
IV. Lesson 18: Radiation
U238 – Po234
1H1
1H2 Deuterium
1H3 Tritium
H2O
2 1H1 = 1H2 + b+
1H2 + 1H1 = 2He3
2He3 + 2He3 = 2He4 + 21H1 + E
Ra 1622 years = ½ Ra; another 1622 years =
¼ Ra, etc.
V. Lab Exercise #14: Radioactivity
VI. Homework Set 5: Hand out, due Thursday, 7/1
VII. Essay 5: Enrico Fermi, due Thursday 7/1
WEDNESDAY, JUNE 30, 2010
I Introduction
II Logistics:
a. Return of papers
b. Running grades
III Review Lesson 17: Spectroscopy
Isaac Newton
Gas hydrogen
_____ n = 2
__e__ n = 1 ground state give off light
IV. Lesson 18: Radiation
U238 – Po234
1H1
1H2 Deuterium
1H3 Tritium
H2O
2 1H1 = 1H2 + b+
1H2 + 1H1 = 2He3
2He3 + 2He3 = 2He4 + 21H1 + E
Ra 1622 years = ½ Ra; another 1622 years =
¼ Ra, etc.
V. Lab Exercise #14: Radioactivity
VI. Homework Set 5: Hand out, due Thursday, 7/1
VII. Essay 5: Enrico Fermi, due Thursday 7/1
LAB 14 RADIOACTIVITY
Physics Lab 14, Wednesday June 30, 2010 Name _____________
Dr Dave Menke, Instructor
I Title: Radioactivity
II Purpose: In this laboratory, students will study radioactive decay. One sample will emit alpha particles. Another, beta particles. Finally, gamma rays will blast from the third sample.
The relationship for radioactive decay is: A = Ao e-t/t
III Equipment
Radiation Detector
The radiation detector is a model Monitor 4/4EC, made by SE International, Inc. Summertown, Tennessee. The units listed are: CPM and mR/hr.
CPM is “Counts per minute,” and it is a measure of radioactive decay. It is the number of atoms within a given radioactive sample that are detected to have decayed in one minute of time.
There is also a term, DPM, which is decays per second. We will use this later.
Seeing “mfR/hr” tells us the number of Roentgens. The Roentgen is a unit of measurement for ionizing radiation, usually for x-rays and g-rays. The unit of 1.0 R is approximately 2 billion units. Also, 1.0 R = 2.58×10−4 C/kg (from 1 esu ≈ 3.33564 × 10−10 C and the standard atmosphere air density of ~1.293 kg/m³)
Also noted on the equipment is the Curie, Ci. The curie is a unit of radioactivity defined as 3.7×1010 decays per second (DPS)
A commonly-used measure of radioactivity is the micro-curie, mCi. In this case, 1.0 μCi = 3.7×104 disintegrations per second = 2.22×106 disintegrations per minute
The typical human body contains roughly 0.1 μCi of naturally occurring something.
Set of 3 samples of radioactive substances:
Sample 1: Polonium-210, emits an Alpha particle, a, at the rate of 0.1 mCi, with a half-life of t = 138.4 days
Polonium-210 decays into two “lighter” elements. One is the Helium nucleus, 2He4. This nucleus is often called an “Alpha” particle, or just using the Greek letter, a. A single gram of 210Po can generate 140 watts of power.
Sample 2: Cobalt-60, emits a Gamma Ray, g, at the rate of 0.1 mCi, and has a half-life of t = 5.27 years.
Due to its short half-life, it is not found in nature. It is created artificially by being bombarded by neutrons, and as a result, Cobalt-60 gives off a beta particle, b-, to become another element which then emits two gamma rays, with energies of 1.17 MeV and 1.33 MeV. Cobalt-60 nuclear bombs leave behind many insidious radioactive isotopes, thus, it is called a “dirty” bomb.
Sample 3: Strontium-90, emits beta particles at the rate of 0.1 mCi, and with a half-life of t = 28.8 years
Strontium-90 is a radioactive isotope which decays to another element while giving off a beta particle, and this new element, in turn, decays quickly (with a t = 64 hours) to become yet another element, giving off more beta particles.
Strontium-90 is a by-product of nuclear fission as fallout from testing nuclear bombs. Along with Cesium-134 and Cesium-137, Strontium-90 were the most important radioactive fallout of the Chernobyl Disaster.
Accidental mixing of radioactive sources containing strontium with metal scrap can result in production of radioactive steel. Discarded radioisotope thermoelectric generators are a major source of 90Sr contamination in the area of the former Soviet Union.
IV Procedure
Turn on Radiation Detector
Sing “Just Dance” by Lady Gaga
Adjust the Detector to get the correct reading of 1x or 10x or 100x.
Use the x-ray glasses to observe your lab partner, who will appear nude
Repeat with Sample 2
Repeat with Sample 3
Determine and write the Chemical Formula for the decay of Sample 1.
Repeat for Sample 2
And for Sample 3
Put away your toys
V Data, Observations, and Calculations go here
VI Results
VII Error
VIII Quests
If you started with 100 grams of Polonium-210 today, how many children would you have in 5 years?
If you started with 100 grams of Cobalt-60 today, how much would remain after 5 years?
If you started with 100 grams of Strontium-90 today, how much would remain after 5 years?
Describe the Chernobyl Disaster
Dr Dave Menke, Instructor
I Title: Radioactivity
II Purpose: In this laboratory, students will study radioactive decay. One sample will emit alpha particles. Another, beta particles. Finally, gamma rays will blast from the third sample.
The relationship for radioactive decay is: A = Ao e-t/t
III Equipment
Radiation Detector
The radiation detector is a model Monitor 4/4EC, made by SE International, Inc. Summertown, Tennessee. The units listed are: CPM and mR/hr.
CPM is “Counts per minute,” and it is a measure of radioactive decay. It is the number of atoms within a given radioactive sample that are detected to have decayed in one minute of time.
There is also a term, DPM, which is decays per second. We will use this later.
Seeing “mfR/hr” tells us the number of Roentgens. The Roentgen is a unit of measurement for ionizing radiation, usually for x-rays and g-rays. The unit of 1.0 R is approximately 2 billion units. Also, 1.0 R = 2.58×10−4 C/kg (from 1 esu ≈ 3.33564 × 10−10 C and the standard atmosphere air density of ~1.293 kg/m³)
Also noted on the equipment is the Curie, Ci. The curie is a unit of radioactivity defined as 3.7×1010 decays per second (DPS)
A commonly-used measure of radioactivity is the micro-curie, mCi. In this case, 1.0 μCi = 3.7×104 disintegrations per second = 2.22×106 disintegrations per minute
The typical human body contains roughly 0.1 μCi of naturally occurring something.
Set of 3 samples of radioactive substances:
Sample 1: Polonium-210, emits an Alpha particle, a, at the rate of 0.1 mCi, with a half-life of t = 138.4 days
Polonium-210 decays into two “lighter” elements. One is the Helium nucleus, 2He4. This nucleus is often called an “Alpha” particle, or just using the Greek letter, a. A single gram of 210Po can generate 140 watts of power.
Sample 2: Cobalt-60, emits a Gamma Ray, g, at the rate of 0.1 mCi, and has a half-life of t = 5.27 years.
Due to its short half-life, it is not found in nature. It is created artificially by being bombarded by neutrons, and as a result, Cobalt-60 gives off a beta particle, b-, to become another element which then emits two gamma rays, with energies of 1.17 MeV and 1.33 MeV. Cobalt-60 nuclear bombs leave behind many insidious radioactive isotopes, thus, it is called a “dirty” bomb.
Sample 3: Strontium-90, emits beta particles at the rate of 0.1 mCi, and with a half-life of t = 28.8 years
Strontium-90 is a radioactive isotope which decays to another element while giving off a beta particle, and this new element, in turn, decays quickly (with a t = 64 hours) to become yet another element, giving off more beta particles.
Strontium-90 is a by-product of nuclear fission as fallout from testing nuclear bombs. Along with Cesium-134 and Cesium-137, Strontium-90 were the most important radioactive fallout of the Chernobyl Disaster.
Accidental mixing of radioactive sources containing strontium with metal scrap can result in production of radioactive steel. Discarded radioisotope thermoelectric generators are a major source of 90Sr contamination in the area of the former Soviet Union.
IV Procedure
Turn on Radiation Detector
Sing “Just Dance” by Lady Gaga
Adjust the Detector to get the correct reading of 1x or 10x or 100x.
Use the x-ray glasses to observe your lab partner, who will appear nude
Repeat with Sample 2
Repeat with Sample 3
Determine and write the Chemical Formula for the decay of Sample 1.
Repeat for Sample 2
And for Sample 3
Put away your toys
V Data, Observations, and Calculations go here
VI Results
VII Error
VIII Quests
If you started with 100 grams of Polonium-210 today, how many children would you have in 5 years?
If you started with 100 grams of Cobalt-60 today, how much would remain after 5 years?
If you started with 100 grams of Strontium-90 today, how much would remain after 5 years?
Describe the Chernobyl Disaster
LAB 13
Physics Lab 13, Tuesday, June 29, 2010 Name _____________
Dr Dave Menke, Instructor
I. Title: Spectroscopy
II. Purpose: to observe the dispersion of light and both bright and dark lines. Each spectroscope is a cylindrical tube, with a slit in one end and a circular diffraction grating at the other end.
III. Equipment: Diffraction Gratings, and/or Student spectroscopes; light sources (incandescent and gas)
IV. Procedure:
1. Take a diffraction grating (and later, a spectroscope) and point it at various light sources
2. Identify each light source used [incandescent light, gas discharge tube – Hg, He, H, Ne, or other element(s)], Sun
3. Observe and record observations. Remember that you must point the end with the slit towards the light source so that the slit is horizontal (left and right, NOT up and down). Look through the diffraction grating end, and notice to the left and to the right of the lighted slit there will be either a “rainbow” of colors, or individual bright lines.
4. Repeat 1 – 3 with a spectroscope
V. Data, observations, calculations:
A. With Diffraction Slide
Observation # Source of Light Observation
1 Light bulb
2 Sun
3 Hydrogen tube
4 Helium
5 Hg (Mercury)
6 Ne
7 Argon
(Expand your table if there are additional light sources available.
B. With Spectroscope
Observation # Source of Light Observation
1 Light bulb
2 Sun
3 Hydrogen tube
4 Helium
5 Hg (Mercury)
6 Ne
7 Argon
(Expand your table if there are additional light sources available.)
VI. Results: (This section is the same as “Conclusions,” so draw your conclusions. Start with reviewing section II Purpose, and determine if you achieved your purpose and at what level of success, e.g., complete failure to phenomenal success, and explain your level)
VII.Error Analysis
A. Quantitative: NA
B. Qualitative:
1. Personal – What did you, or your partner(s) do to screw up?
2. Systematic – What part of the environment screwed things up, e.g., weather, equipment, etc.
3. Random – As this is an observational experiment without doing multiple trials and taking averages, there is random error. We don’t know what it is, thus, we call it “random.”
VIII. Questions:
A. What is a continuous spectrum, and what causes it?
B. What is a bright line spectrum?
C. What is a dark line spectrum?
D. What kind of spectrum does the Sun have, and why?
Dr Dave Menke, Instructor
I. Title: Spectroscopy
II. Purpose: to observe the dispersion of light and both bright and dark lines. Each spectroscope is a cylindrical tube, with a slit in one end and a circular diffraction grating at the other end.
III. Equipment: Diffraction Gratings, and/or Student spectroscopes; light sources (incandescent and gas)
IV. Procedure:
1. Take a diffraction grating (and later, a spectroscope) and point it at various light sources
2. Identify each light source used [incandescent light, gas discharge tube – Hg, He, H, Ne, or other element(s)], Sun
3. Observe and record observations. Remember that you must point the end with the slit towards the light source so that the slit is horizontal (left and right, NOT up and down). Look through the diffraction grating end, and notice to the left and to the right of the lighted slit there will be either a “rainbow” of colors, or individual bright lines.
4. Repeat 1 – 3 with a spectroscope
V. Data, observations, calculations:
A. With Diffraction Slide
Observation # Source of Light Observation
1 Light bulb
2 Sun
3 Hydrogen tube
4 Helium
5 Hg (Mercury)
6 Ne
7 Argon
(Expand your table if there are additional light sources available.
B. With Spectroscope
Observation # Source of Light Observation
1 Light bulb
2 Sun
3 Hydrogen tube
4 Helium
5 Hg (Mercury)
6 Ne
7 Argon
(Expand your table if there are additional light sources available.)
VI. Results: (This section is the same as “Conclusions,” so draw your conclusions. Start with reviewing section II Purpose, and determine if you achieved your purpose and at what level of success, e.g., complete failure to phenomenal success, and explain your level)
VII.Error Analysis
A. Quantitative: NA
B. Qualitative:
1. Personal – What did you, or your partner(s) do to screw up?
2. Systematic – What part of the environment screwed things up, e.g., weather, equipment, etc.
3. Random – As this is an observational experiment without doing multiple trials and taking averages, there is random error. We don’t know what it is, thus, we call it “random.”
VIII. Questions:
A. What is a continuous spectrum, and what causes it?
B. What is a bright line spectrum?
C. What is a dark line spectrum?
D. What kind of spectrum does the Sun have, and why?
LESSON 17
PHYSICS LESSON 17 FOR
TUESDAY, JUNE 29, 2010
I Introduction
II Logistics:
a. Return of papers
b. Running grades
III Review Lesson 16: Sound Waves, Light Waves
IV. Lesson 17: Spectroscopy
Isaac Newton
Gas hydrogen
_____ n = 2
__e__ n = 1 ground state give off light
V. Lab Exercise #13: Spectroscopy
VI. Homework Set 5: Hand out, due Thursday, 7/1
VII. Essay 5: Enrico Fermi, due Thursday 7/1
TUESDAY, JUNE 29, 2010
I Introduction
II Logistics:
a. Return of papers
b. Running grades
III Review Lesson 16: Sound Waves, Light Waves
IV. Lesson 17: Spectroscopy
Isaac Newton
Gas hydrogen
_____ n = 2
__e__ n = 1 ground state give off light
V. Lab Exercise #13: Spectroscopy
VI. Homework Set 5: Hand out, due Thursday, 7/1
VII. Essay 5: Enrico Fermi, due Thursday 7/1
Monday, June 28, 2010
LAB 12 Thin Films
Physics Lab 12 June 28, 2010 Name __________________
I. Title: Thin Films
II. Purpose: to observe the effects of thin films on destructive interference of light, and to determine the mass of an average bubble
III. Equipment: Soapy fluid (source of bubbles) - individual bottles of bubbles; a wand to make the bubbles; a blower; a digital scale; stopwatch
IV. Procedure:
1. Have a lab partner acquire the stuff needed.
2. Weigh an individual closed bottle of bubbles, with wand inside, in grams. Record.
3. Choose a lab partner to be the “blower.”
4. Have the “blower” carefully remove the “wand” to extract fluid, then have the blower create bubbles by blowing; don’t drop fluid onto floor or table, as that will create personal error
5. While the blower is “blowing,” have another partner count the number of bubbles. Have yet another partner determine how long it takes for an average bubble to exist before popping. Finally, have an additional person note the different colors on the bubble’s “skin” as it slowly vanishes.
6. Blow about 100 bubbles. Do not confuse your group’s bubbles with bubbles blown by other blowers in other groups.
7. As the bubbles are being created and counted, carefully observe a number of bubbles as they become invisible at the top before they pop. Record observations. This will be an example of thin film destructive interference.
8. When completed, return the wand to the bottle and seal the bottle.
9. Weigh the cylinder again in grams. Record.
10. Subtract the weight of the bottle in #9 from the weight of the bottle in #2. Record. (This is the mass of the fluid used to make bubbles).
11. Calculate the average mass of a bubble by dividing the mass you found in #10 by 100 (or as many bubbles as you counted). Record.
V. Data, observations, calculations:
1. Mass of sealed bubble bottle: ________ grams
2. Number of bubbles blown: _________
3. Mass of bottle after bubbles are blown:_____ g
4. Mass of soapy fluid used: _______ g
5. Average mass of a bubble: _______ g
6. Observations of the thin films:
a. Colors
b. “life” of a bubble: _______ seconds
c. invisibility
VI. Results: As ever, re-read the purpose (Section II) and determine if you achieved this purpose, and how well you achieved it, i.e., what are your conclusions?
VII. Error:
Quantitative: NA
Qualitative:
1. Personal – what did you or your partner(s) do to screw up?
2. Systematic – any equipment failures or violent thunderstorms in the lab while doing this experiment?
3. Random – always random error, unless you do multiple trials and then take an average …
VIII. Questions:
1. How much time did the average bubble last before it broke?
2. What colors did you notice in the bubbles?
3. Why did some of the bubbles rise in the air, rather than fall?
4. How would this lab be different if the room temperature were much colder?
I. Title: Thin Films
II. Purpose: to observe the effects of thin films on destructive interference of light, and to determine the mass of an average bubble
III. Equipment: Soapy fluid (source of bubbles) - individual bottles of bubbles; a wand to make the bubbles; a blower; a digital scale; stopwatch
IV. Procedure:
1. Have a lab partner acquire the stuff needed.
2. Weigh an individual closed bottle of bubbles, with wand inside, in grams. Record.
3. Choose a lab partner to be the “blower.”
4. Have the “blower” carefully remove the “wand” to extract fluid, then have the blower create bubbles by blowing; don’t drop fluid onto floor or table, as that will create personal error
5. While the blower is “blowing,” have another partner count the number of bubbles. Have yet another partner determine how long it takes for an average bubble to exist before popping. Finally, have an additional person note the different colors on the bubble’s “skin” as it slowly vanishes.
6. Blow about 100 bubbles. Do not confuse your group’s bubbles with bubbles blown by other blowers in other groups.
7. As the bubbles are being created and counted, carefully observe a number of bubbles as they become invisible at the top before they pop. Record observations. This will be an example of thin film destructive interference.
8. When completed, return the wand to the bottle and seal the bottle.
9. Weigh the cylinder again in grams. Record.
10. Subtract the weight of the bottle in #9 from the weight of the bottle in #2. Record. (This is the mass of the fluid used to make bubbles).
11. Calculate the average mass of a bubble by dividing the mass you found in #10 by 100 (or as many bubbles as you counted). Record.
V. Data, observations, calculations:
1. Mass of sealed bubble bottle: ________ grams
2. Number of bubbles blown: _________
3. Mass of bottle after bubbles are blown:_____ g
4. Mass of soapy fluid used: _______ g
5. Average mass of a bubble: _______ g
6. Observations of the thin films:
a. Colors
b. “life” of a bubble: _______ seconds
c. invisibility
VI. Results: As ever, re-read the purpose (Section II) and determine if you achieved this purpose, and how well you achieved it, i.e., what are your conclusions?
VII. Error:
Quantitative: NA
Qualitative:
1. Personal – what did you or your partner(s) do to screw up?
2. Systematic – any equipment failures or violent thunderstorms in the lab while doing this experiment?
3. Random – always random error, unless you do multiple trials and then take an average …
VIII. Questions:
1. How much time did the average bubble last before it broke?
2. What colors did you notice in the bubbles?
3. Why did some of the bubbles rise in the air, rather than fall?
4. How would this lab be different if the room temperature were much colder?
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