Students

I read your letters telling me why you did not do your homework to find the energy and frequency of light at different wavelengths . Many of you complained that you did not know what to do. I won’t argue with you. I am going to teach you exactly what to do, step-by-step.

1. What this is all about -- There is an important relation between the wavelength of light, its frequency and the energy it contains. Lets make sure you remember what wavelength, frequency and energy mean.

a. wavelength – When you toss a stone in a pool of water, you see ripples, or “waves.” The waves are energy being sent from the spot where the stone hit the water to a place were the energy is released. The distance from the top, or crest, of one wave to the top of the next is the length of the wave. Water, sound, earthquakes, radio and light are just some sources of waves. Water waves need oceans, rivers, pools, etc., to travel. Sound requires air, liquid or solid. It will not travel through a vacuum. Earthquake waves travel through the earth. Light and radio waves are forms of “electromagnetic radiation.” They can travel through the vacuum of space.

Radio waves are usually measured in meters or even kilometers. Light is much, much shorter. We measure the wavelength of light in units called “nanometers.” A nanometer, abbreviated “nm,” is one billionth of a meter. In other words, it would take one billion (1,000,0000,000) nanometers lined up end to end to equal one meter.

b. frequency – The number of times something happens is called “frequency.” The earth rotates on its axis one time each 24 hours. It has a frequency of one time per day. Electricity produced by generators changes its direction as it is produced. In the United States, it changes direction 60 times per second. Each complete change of direction is a “cycle.” We say that electricity has a frequency of 60 cycles per second. Sound, light and radio waves, among others, also are cyclic. This means that one wave follows another. (If you have just one wave, it is a “pulse.”) The number of waves that pass a given point in some unit of time is the frequency of the wave. If the unit of time is one second, then the frequency is given the special name “Hertz.”

It is very difficult to count the waves as they pass so we find frequency another way. We know that waves move at a constant speed through whatever they are traveling. When light or radio waves move through a vacuum, they travel at 299,800,000 meters per second (m/sec). This is the “speed of light.” If we divide the speed of light by the length of light or radio waves, the answer is the number of waves passing in a second.

In symbols we would write f = c / l. . The symbol “c” stands for the speed of light. The symbol “l” (Greek letter “lambda”) stands for the wavelength. Lets assume that the wavelength is one meter (1 m). This is a billion times longer than a nanometer, so it has to be a radio wave, not a light wave. To find the frequency, we divide 299,800,000 m/sec by 1 m.

The meter in the numerator cancels the meter in the denominator. Then the frequency is 299,800,000 / s . We read this as “frequency is two hundred ninety nine million eight hundred thousand per second.”

c. energy – Radio and light waves are called “electromagnetic radiation” and occupy different parts of the “electromagnetic spectrum.” They carry energy and the amount of energy depends upon the frequency of the wave. Max Planck, a great physicist, studied the relationship between frequency and the energy of the wave. He found that the relationship was constant. In other words, dividing the energy of a wave at a specific wavelength by the frequency at that wavelength always, always gave the same result: 6.626068 × 10^-34 m² kg / s. That is h = E / f.

In the equation,”h” is the constant, named “Planck’s constant” in honor of Max Planck. “E” is the energy and “f” is the frequency. We usually rearrange the equation to solve for Energy: E = hf.

Notice two things about the constant. First, it is written in scientific notation. This is because the number is so small. If we write it out, it would be 0.0000000000000000000000000000000006626068 m²kg / s. In this form, the number is very difficult to work with. Second, the unit of energy “m²kg / s” is read as “meter squared kilogram per second.” It is a strange unit and we will talk about it in a little while.

2. Energy and electron orbitals -- Neils Bohr studied how the orbitals occupied by electrons in atoms were related to their energy. His work gave us what we call the “Bohr model of the atom.” It is very important. Bohr found that that the light emitted by electrically excited hydrogen related to the orbitals that hydrogen’s only electron could occupy. Look at the spectrum of light emitted by hydrogen:

To find the frequency at each wavelength we divide the wavelength into the speed of light. We have to use a “conversion factor” to change nanometers to meters so our units are correct. As an example we will find the frequency for the wavelength of 656 nm.

We will write the numbers in scientific notation format. We can do this because our number system is based on “10.” That is, 100 can be written as 1 x 10 x 10 or 1 x 10², 200 as 2 x 10 x 10 or 2 x 10², 1000 as 1 x 10 x 10 x 10 or 1 x 10³ and so on. That is great for numbers 1 or more. What about numbers less than 1? For example, what is 1/100? The result by long division is 0.01 and we read it as “one hundredth.” In scientific format we move the decimal point to the right until we have one non-zero digit to the left of the decimal. In this case that is two decimal places. We write the number as 1 x 10^-2. The “-“ in the exponent indicates that the number is less than one. Then in our problem, we get:

Notice in the term (1 m /1 x 10⁹ nm), “10⁹ nm” is in the

denominator of the denominator[1]. Using powers of 10 makes the math easier. Use the rule that to divide powers of 10, you subtract exponents. To multiply, you add exponents. We divided 6.56 into 2.998 by long division. Our answer is
f = 4.57 x 10¹⁴/s because we apply the rule of one digit to the left of the decimal point. Do you see? We have cancelled like units in the numerator and denominator. We read the answer, as “The frequency is 4.57 x 10¹⁴ per second.”

To find energy, E, multiply 4.57 x 10¹⁴/s times 6.626068 × 10^-34 m² kg/s. The answer is 30.281 x 10 ^-20 m² kg/s². The unit “m² kg/s²” is called a “Joule,” named after English physicist James Prescott Joule (1818–89).

Students, please find the energy for the other 3 wavelengths of visible light in hydrogen’s emission spectrum. Then prepare a graph plotting the wavelength, l, against the Energy in Joules. For which wavelength is the energy least? For which is it greatest? What are the associated colors? Please see me for questions. When you have completed the graph, we will conduct the “glowing star” experiment.

JAY L. STERN, 26 March 2011

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[1] de·nom·i·na·tor noun \di-ˈnä-mə-ˌnā-tər\ the part of a fraction that is below the line and that functions as the divisor of the numerator

Today is 2 February 2011 -- Today we are studying electric circuits

How many kinds of electric circuits do you think there are? To know how to answer this question means knowing what an electric circuit is. Begin by examining the word “circuit:”

cir·cuit   

[sur-kit]–noun

1. an act or instance of going or moving around.

2. a circular journey or one beginning and ending at the same place; a round.

3. a roundabout journey or course.

4. a periodical journey from place to place, to perform certain duties, as by judges to hold court, ministers to preach, or salespeople covering a route.

5. the persons making such a journey.

6. the route followed, places visited, or district covered by such a journey.

7. the line going around or bounding any area or object; the distance about an area or object.

8. the space within a bounding line; district: the circuit of the valley.

Do you see what these definitions have in common? If you said “goes around,” you are correct. A circuit involves leaving a point, going on a trip along a defined course, and returning to the original point.

Now, lets be a little more specific. What is an “electric circuit?” Then we have just one definition, but it has two parts:

Electricity .

a. Also called electric circuit. the complete path of an electric current, including the generating apparatus, intervening resistors, or capacitors.

b. any well-defined segment of a complete circuit.

So now you know what an electric circuit is. The apparatus, equipment or device that supplies the electricity can be a generator, a battery, a capacitor or even static electricity. In other words, anything that causes electrons to flow from one point to another will drive an electric circuit. A resistor is – strangely enough – anything that resists the flow of electricity as it goes around the circuit. It can be an electric heater, a light bulb, a motor or anything that draws energy from the flow of electrons and converts it into work, heat, light or other form of energy. A capacitor stores electrons and can discharge them upon demand.

What the definition failed to include is the fact that a conductor is necessary, too. A conductor is anything that allows the electrons to flow. It usually is a metal wire – metals conduct electricity. But it can also be a pool of salt water, a carbon rod or charged particles such as anions and cations. (We will talk about these at another time.)

To keep the metal wire from conducting electricity to things we don’t want to be involved in our circuit, we need an insulator. This is something coating the wire that does not conduct electricity. Rubber is a great insulator and it is commonly used.

There are just two types of electric circuits: a series circuit and a parallel circuit. The series circuit means that two or more devices are connected – or “wired” – together one after the other. The parallel circuit means that the two or more devices are connected to the source of electricity independently. Look at the illustrations. Which one is which? Point out the series diagram and the parallel circuit.

By the way, notice some of the symbols used. The pairs of short/long parallel lines represent a battery. The “+” is the positive terminal, or cathode. The “-“ is the negative terminal, or anode. In a circuit, electrons flow through the conductor from the anode to the cathode.

After you understand the fundamentals of electric circuits, you are armed with knowledge. For example, you know that if an insulator is placed in the path of a circuit, electrons will not flow. You may observe it, in which case we call it a conclusion. Or you may hypothesize that no electrons will flow. Then we say you are basing your idea on research about electrical phenomena that other people have conducted. If you actually observe how the circuit works, then you are producing a theory of your own.

Today is 25 January 2011 -- We learned about Mendelian Genetics

Today we learned about segregation, independent assortment and the concept by Gregor Mendel that two "elements" contributed to inheritance of characteristics. His work was long before scientists understood chromosomes and genes, and the discovery of DNA (deoxyribosenucleic acid). Be sure to understand the terms Mendel pioneered. What are: F1, F2, P1, recessive, dominant? How did statistical analysis of the results of Mendel's experiments (he used pea plants grown in the monastery garden as test subjects) lead him to formulate his laws of genetics? Why is this useful today?

Today is 19 January 2011 -- How long does sound take to travel?

Today is 19 January 2011 -- How long does sound take to travel?

The volcano Krakatoa erupted 27 August 1883 with a sound so loud it was heard 3,000 miles away, on Rodriguez Island. You know the speed of sound (It was given in one of the video presentations as 1100 ft/sec.) Although the speed of sound changes based on air density and temperature, we can assume the speed to be the same as under normal atmospheric conditions. How do you solve this?

The equation is still S = r x t, where S = distance, r = speed or velocity, and t = time. Rearrange the equation to solve for time: S/r = t. Then substitute the known information: 3,000 mi/1100 ft per sec = t. You need to convert miles to feet or feet to miles to have the same units. Since students were taught there are 5,280 feet per mile, we can write:

[(3,000 miles)(5280 feet/mile)] / (1100 feet/sec) = time.

In class I taught the students to use scientific notation to solve the problem because it is easier. I also reminded them to cancel units where possible and to remember that the denominator "goes into the numerator," not the other way around. Here are the steps:

[(3,000 miles)(5280 feet/mile)] / (1100 feet/sec) = time.

(3.0 x 10³ mile)(5.28 x 10³ feet /mile) 15.84 x 10⁶

-------------------------------------------- = --------------

(1.1 x 10³ feet/sec) 1.1 x 10³/sec

Divide 15.84 x 10⁶ (numerator) by 1.1 x 10³ (denominator). Work first with the real numbers:

_____

1.1 ) 15.84

Move the decimal point of the divisor (1.1) over to the right one place until it is a whole number. What you do “outside,” you have to do “inside” too:

_____

11) 158.4

Now, ask yourself, “Does 11 go into 1?” No. So then ask, “Does 11 go into 15?” Yes, once. So write a “1” on the line over the “5,” multiply 1 x 11, write the answer below the 15, then subtract from 15:

__1___

11) 158.4

11

4

Then, bring down the “8” and ask how many times 11 goes into 48. It goes 4 times, so you then multiply 11 x 4 on the next line and write “44.” Subtract 44 from 48 to get “4.”

__14___

11) 158.4

11

48

44

4

Bring down the last digit, “4” and divide 11 into another 44:

__14.4___

11) 158.4

11

48

44

44

44

You have no remainder so you are finished. Remember to place a decimal point above the one inside the box.

Next, handle the exponential term. When you multiply exponentials, you add them; when you divide, you subtract:

10⁶ / 10³ means 10^{(6 – 3)} = 10³

So your answer is 14.4 x 10³ seconds. How did the unit “seconds” get into the numerator? It is because it was the denominator of a denominator. When you divide, you invert the denominator and multiply, which puts the seconds into the numerator.

Finally, how many seconds are there in one hour? There are 3600 or in scientific format, 3.6 x 10³ sec/hr. As we did before, divide 3.6 into 14.4. What is your answer? I got “4 hours.” Remember, the exponential terms will cancel out as will the unit “seconds.”

The people on Rodriguez Island reported hearing sounds like the “roar of canons” on the night of 26 August 1883. It took 4 hours for the sound of Krakatoa erupting to reach them. By the way, how can it be that the eruption happened on 27 August and they heard it the night of the 26^th?

Today is 14 January 2011 -- We had a test on sound waves and to write "Press On" by Calvin Coolidge

Here is the test:

1. Write -- from memory and word-for-word -- the inspirational quotation “Press On” by the 30^th president of the United States, Calvin Coolidge:

"Nothing in the world can take the place of persistence. Talent will not; the world is full of unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not; the world is full of educated derelicts. Persistence and determination alone are omnipotent. The slogan “press on” has solved and always will solve the problems of the human race. "

2. Look at the equation f = v / l.

a. What does each symbol stand for?
f = frequency (times something happens per unit of time)
v = velocity (distance per unit of time, like “meters per second”)
l = wavelength

b. When or how is this equation used?

Find wavelength when frequency and wave velocity are known; find frequency when wavelength and velocity are known; find velocity when wavelength and frequency are known.

c. Compare this equation to the equation relating distance to speed and time at that speed,
S = r x t. (Which term in the distance equation corresponds to which term in the unknown equation?)

S is distance and compares to wavelength

r is rate in units of distance per unit of time and compares to velocity

t is time and relates to frequency which is the number of cycles per unit of time.

3. Answer the following definition questions:
a. What travels at 186,000 miles per second?

Light, electromagnetic radiation in genera

b. How many feet in a mile?

5280 feet per mile

c. How many centimeters in one inch?

2.54 cm per inch

4. A tube closed at one end is 13 cm long. Blowing across the open end of the tube makes a standing sound wave.
a. How much of the wave is in the tube?

¼ of wave is in closed-end tube.

b. What is the length of the full wave?

Since ¼ fits in tube, full wave must be 4 x 13cm = 52 cm.

c. The frequency of the wave is 644.77 per second. Calculate the speed of sound for this condition.
v = f x l = 644.77/sec x 52 cm = 33,528 cm / sec

d. The period of a wave is the reciprocal of its frequency, P = 1/f. What is the period of the wave?

P = 1 / 644.77 / sec = 0.00155 seconds
5. Are sound waves longitudinal or transverse? How can you show this?
Longitudinal. They are compressive which gives rise to the Doppler effect.

6. “The cheetah is capable of speed up to 72 mph (114 km/h) and can maintain this speed over an average prey chase of 3.5 miles."
a. How long will the cheetah run at this speed?
Use equation for distance: S = r x t. Rearrange to solve for time: S / r = t = 3.5 mile/72 mile/hr.

= 0.0486 hr. For minutes, multiply by 60 min/hr for 2.91 minutes.

b. If a marathon runner can go 15 mph (for 26 miles), what minimum distance ahead of the cheetah does he have to be to keep from being the cheetah’s lunch? Assume that the cheetah stops running exactly at 3.5 miles. (This problem is a little tricky but you can do it if you THINK!)

The runner has to be far enough ahead so that the cheetah can’t catch him in 2.91 minutes. In that time, the runner can go S = r x t = 15 m / hr x 0.0486 hr = 0.729 mile. That means he has to be 3.5 – 0.729 miles or 2.771 miles.

c. You will be relieved to know that the marathoner escaped the cheetah! If his home village is the same distance as his minimum distance, and he shouts very loud “I’m safe!” how long will the sound of his voice take to reach the village?

Since sound travels 1100 ft / sec, and there are 5280 feet per mile, the time it takes the sound to reach the village is (2.771 miles x 5280 feet / mile) / 1100 ft/sec = 13.3 seconds.

Today is 15 December 2010 -- We are learning a little about cell growth and reproduction

Students had three questions to answer to see if they were paying attention. Here are the questions and a sampling of the answers:

1. How do metastases affect an individual with cancer?

"They can be convinced in other parts of the body."

"It spreads diseased cells to other parts of the body by way of blood or lymphatic vessels or membranous surfaces."

"The break-away cells can be carried in other parts of the body."

"It spreads faster."

"Allows the cancer cells to grow at a specific rate."

2. What does the "a" in asexual mean?

"One individual person."

"Single - erotic"

"Single - not having a partner."

"The 'a' in asexual refers to the asexual reproduction which needs only a single individual."

"No partner with you."

"It means you can create cancer."

"Means sex by yourself -- masturbation."

3. What is the difference between "benign" and "malignant?"

"I don't know."

"That they mean 2 different things and have different meanings."

"They are opposite."

"Benign is the cancer cells typically (grouped) together; malignant is a break-away cell."

"Benign is a growth that is not harmful and malignant is the opposite."

Most of the answers showed that the students had at least a passing acquaintance with the subject. Of course, a few students will be, I'm afraid, rather disappointed with the true meaning of "asexual." It means to reproduce the organism without the benefit of an exchange of genetic material with another individual. In this respect, the offspring is a clone of the parent.

Students were also assigned review questions 1, 2 and 3 on pages 137 and 142 of their text. The questions dealt with an understanding of the cell cycle (p. 137) and of mitosis (p. 142).

Bravo to those students who completed the homework and who really paid attention to details of the cell cycle!

Today is 11 January 2011 -- What do we need to know about Genetics?

Here are the California Standards for Genetics. We will not be studying each and every detail, but this gives you an idea of what topics there are within the subject area that students throughout California learn about.

2. Mutation and sexual reproduction lead to genetic variation in a population. As a basis for understanding this concept:

a. Students know meiosis is an early step in sexual reproduction in which the pairs of chromosomes separate and segregate randomly during cell division to produce gametes containing one chromosome of each type.

b. Students know only certain cells in a multicellular organism undergo meiosis.

c. Students know how random chromosome segregation explains the probabilitythat a particular allele will be in a gamete.

d. Students know new combinations of alleles may be generated in a zygote through the fusion of male and female gametes (fertilization).

e. Students know why approximately half of an individual’s DNA sequence comes from each parent.

f. Students know the role of chromosomes in determining an individual’s sex.

g. Students know how to predict possible combinations of alleles in a zygote from the genetic makeup of the parents.

3. A multicellular organism develops from a single zygote, and its phenotype depends on its genotype, which is established at fertilization. As a basis for understanding this concept:

a. Students know how to predict the probable outcome of phenotypes in a genetic cross from the genotypes of the parents and mode of inheritance (autosomal or X-linked, dominant or recessive).

b. Students know the genetic basis for Mendel’s laws of segregation and independent assortment.

c.* Students know how to predict the probable mode of inheritance from a pedigree diagram showing phenotypes.

d.* Students know how to use data on frequency of recombination at meiosis to estimate genetic distances between loci and to interpret genetic maps of chromosomes.

4. Genes are a set of instructions encoded in the DNA sequence of each organism that specify the sequence of amino acids in proteins characteristic of that organism. As a basis for understanding this concept:

a. Students know the general pathway by which ribosomes synthesize proteins, using tRNAs to translate genetic information in mRNA.

b. Students know how to apply the genetic coding rules to predict the sequence of amino acids from a sequence of codons in RNA.

c. Students know how mutations in the DNA sequence of a gene may or may not affect the expression of the gene or the sequence of amino acids in an encoded protein.

d. Students know specialization of cells in multicellular organisms is usually due to different patterns of gene expression rather than to differences of the genes themselves.

e. Students know proteins can differ from one another in the number and sequence of amino acids.

f.* Students know why proteins having different amino acid sequences typically have different shapes and chemical properties.

5. The genetic composition of cells can be altered by incorporation of exogenous DNA into the cells. As a basis for understanding this concept:

a. Students know the general structures and functions of DNA, RNA, and protein.

b. Students know how to apply base-pairing rules to explain precise copying of DNA during semiconservative replication and transcription of information from DNA into mRNA.

c. Students know how genetic engineering (biotechnology) is used to produce novel biomedical and agricultural products.

d.* Students know how basic DNA technology (restriction digestion by endonucleases, gel electrophoresis, ligation, and transformation) is used to construct recombinant DNA molecules.

e.* Students know how exogenous DNA can be inserted into bacterial cells to alter their genetic makeup and support expression of new protein products.

Today is 11 January 2011 -- Sound wavelengths and frequencies

Its fun to see how science can be demonstrated with ordinary things. Today we explored sound frequencies using test tubes (or bottles), a meter stick, a candle and a home-made air cannon. This is what we did:

1. Air cannon -- For 99 cents I bought a 2-quart water bottle. I cut off the bottom and got a large balloon. I tied off the mouth of the balloon, then cut off the part of the closed end. I s--t--r--e--t--c--h--e--d the balloon over the bottom of the bottle. I should mention that in cutting off the bottle bottom, I left the rounded part so the balloon would not be punctured by sharp, plastic edges. To keep the balloon from slipping off the bottle, I wrapped the joint between balloon and bottle with electrical tape. Then, when I pulled on the knot where the mouth of the balloon had been, and released it, a "ball" of air shot out of the bottle mouth.

2. Flute -- We used test tubes. The ones on hand were 9.5 cm long, about 1.5 cm in diameter. By blowing over the open mouth of the test tube, we caused a pressure difference which made the tube vibrate. This vibration caused sound of a specific wave length. When we put water in the test tube, we changed its length. As the effective length of the tube shortened, what happened to the pitch of the sound? Did the loudness of the sound depend on the pitch or how hard the students blew across the tube? What happened if the tube was blown into instead of across the mouth? For more information, go to the link "Standing waves and Wind Instruments". There is more information than discussed in our text book (Activity 3, "Sounds from Vibrating Air."

3. Frequency -- We can find the frequency of the sound from the vibrating column of air for any length of tube. The equation is f = v / λ where f is the frequency in cycles per second, v is the velocity of the air and λ is the wavelength. We know that in a tube closed at one end, the wavelength of a vibrating column of air is four times the length of the tube. Another way of saying this is that 1/4 the wavelength fits in the column. If we can determine the velocity of sound in air, we can find the frequency of the sound.

There are two ways we can look at the velocity. One way is "fun," but maybe not so accurate. The other way is by scientific measurement. Lets say you are in a race. On your mark, set, GO! The starter fires his pistol. First you see the puff of smoke from the pistol, then you hear the sound of gunfire. You see the puff of smoke first because light travels so much faster than sound. In normal, still air, sound travels 1100 feet per second. You can substitute the values for the wavelength and the velocity in the equation, above. When you do so, you get: f = (1100 ft /sec x 30.48 cm / ft) / (9.5 cm X 4) = 882 per second.

O.K. Now, lets lay a meter stick on the table. We put a candle at "0 cm" and the air cannon at 100 cm. Students timed how long it took the pulse of air to travel 100 cm and cause the candle flame to flicker or be extinguished. Use the equation above to find out the frequency this way. Which answer makes more "sense" to you? Why?