`Physics 5/ History 177                     Fall 1997/ Spring 1998`

The Nuclear Age: Its Physics and History

Prof. Gary R. Goldstein, Prof. Martin Sherwin

Some Background: Estimation

Physics is a quantitative science and so any appreciation of its contribution to the modern world view obliges us to do some mathematics. We will use some arithmetic, algebra, geometry, and "scientific notation". Don't panic, you'll be surprised at how little is necessary.

The first thing we are going to talk about is estimation of numbers. The ability to estimate the size or probability of various phenomena is useful in many realms of activity beyond science. It is worth developing for the sake of having a better grasp of economic and political decision making in the public and personal spheres as well.

We begin with some estimations that, to the uninitiated, seem totally impossible to accomplish with any confidence. You'll be surprised!

The solutions presented below are not necessarily those worked out in your class, but were inspired by previous classes. Notice that the results are not very different from those obtained in your class although the assumptions and procedures may be quite different. This shows how these estimates are quite insensitive to such details.

1) How many hairs are on a head?

Solution: Where do you begin to answer such a question? Whose head is being discussed? Well, to begin with, since every head is different (from the thickest head of hair to the totally bald) it must be assumed that we're going to take some average person's hair. The method we'll use is quite straightforward. Namely, we will estimate the area of the scalp and then multiply by the estimated number of hairs in a small unit of area.

First, to get the area of the scalp we use a geometric approximation which will be our model. We do this because the actual determination of that area would be difficult to make for any single head and because an exact result (say to 2 figure accuracy) would be representative of only that one head. Since heads vary in size even among adults (probably by 10% to 20%) the model will do as well for the approximation we're trying to obtain. Now the model is a hemisphere of the same diameter as a typical adult head. That diameter is about 10 inches or 25 centimeters. Check your own head by looking in a mirror with a ruler across your face.

The surface area of a hemisphere of radius r is half that of a sphere,

Area of hemisphere = 1/2 area of sphere = 1/2 x 4pr2 = 2pr2.

With the radius, r = 1/2 diameter = 5 in and p @ 3 , we get

Area of "scalp" = 2x3x(5in)2 = 6x25 in2 = 150 in2.

Note that p @ 3 is a close enough approximation to 3.141596... for our purposes; we're aiming for "order of magnitude" accuracy. That means the nearest power of ten, as you'll see.

Next we need some estimate of how many hairs are in a typical patch or unit of area. By counting the number of hairs roughly along a line of 1 inch, somewhere in the middle of your or your friend's scalp, you will get something like 15 to 40. Then squaring that, which assumes the hair is arranged somewhat like a checkerboard or grid, gives 225 to 1600 per square inch.

Combining these two factors we have the number of hairs equals

150 in2 x (200 or 1600)in-2 = 30,000 or 240,000 hairs.

Note that I've rounded off the 225 to 200 which is justified by the large uncertainty in our estimates. The result is somewhere between 0.3 X 105 and 2.4 X 105 (using "scientific notation" for the powers of ten involved). We can say that the relevant order of magnitude is 105.

Then that rough approximation to the answer gives the interesting result that among 10 people there are about one million hairs on their collective heads (excluding beards). So, if you now are told that your chance of winning something (like a lottery for example) is one in a million, you can visualize that. It is the chance of randomly picking one single secretly marked hair in one try from the heads of 10 people. Not very likely!

As a second application, we will see in due course that the explosive power of nuclear warheads is about a million times that of conventional chemical explosives. The realization by European and US physicists that nuclear processes could lead to the release of a million times more energy than chemical (actually molecular) processes was an immediate consequence of the discovery of nuclear fission by the German physicists Hahn and Strassmann and their exiled colleagues, Meitner and Frisch. But one single nucleus undergoing fission releases only about one-millionth of the energy released by a measly 1 Watt light bulb in thirty-millionths of one second! So what was the alarm about? We'll see.

2) How many grains of sand are there on all the beaches of the world?

Solution: We'll simply try to estimate how much sand is in a small volume, then estimate the volume of all the sandy beaches, and multiply. How do we get the volume? Our experience with beaches teaches us what the width of a typical beach is. The depth requires a less certain remembrance of digging down into the sand to reach soil or rock. A more difficult estimation is the length.

a) Estimate the number of sand grains in a cubic centimeter;

One student found that about 20 grains of sand (from a nearby beach) line up along a cm, so

20 x 20 x 20 = 8000 grains/cm3 or about 104 cm-3 .

b) The width of a typical beach is about

100 m = 104 cm.

c) Estimate the depth of a typical beach;

5 to 10 m, say 10 m = 103 cm .

d) If we assume there are sandy beaches along most of the continental coasts in the world then we can imagine a coastal strip of beach circling the earth about 10 times. The circumference of the earth is 2 x p x radius. The radius is approximately 4000 miles or 6000 Kilometers or 6 million meters or 6x106 m. The circumference is thus

2pr @ 2x3x6x106 = 36x106 @ 4x107 m.

10 times that length is 4x108 m = 4x1010 cm .

Then the volume of sand on all the beaches is very approximately

4x1010 x 104 x 103 = 4x1017 cm3.

e) How many grains of sand is this?

Multiply the density of sand grains or the number of grains per cubic centimeter in (a) by the volume we have just obtained in (d):

104 cm-3 x 4x1017 cm3 = 4x1021 grains of sand.

If we overestimated the length by a factor of 4 we would have obtained 1021. If we underestimated the length by not taking inland seas into account properly, say by a factor of 5, we'd have obtained 2x1022. Other approximations as well could raise or lower this. Beach depth could be 2 or 3 times more for example. An order of magnitude estimate would be anywhere from 1020 to 1024 grains of sand. That latter number is roughly the number of molecules in a cubic foot of air!!! That is a large number!

To be specific, there are 6.02x1023 molecules of a gas in 22.4 liters at 0oC and Standard Atmospheric Pressure. That is called Avogadro's Number or No. (At 20oC, room temperature, the same number occupy a slightly larger volume by about 8%.) So you now have some intuitive sense about such an enormous number.

That such an enormous number should exist was conjectured by Avogadro way back in 1811 based on atomic ideas that were then being developed by physicists and chemists. An important inference from Avogadro's conjecture concerns pure samples of single molecules (or atoms) of atomic weight A: there are 6.02x1023 molecules (or atoms) in A grams of the pure substance. The A grams amount (whether solid, liquid or gas) is called a mole.

So 1 mole of a pure substance contains No molecules (or atoms) of that substance.

A relevant example is Uranium, which is an element whose atomic weight is approximately 238. Then about 240 grams or 1/4 Kg or 1/2 lb of pure Uranium contains 6.02x1023 atoms.

Now if ordinary matter contains such huge numbers of molecules or atoms or nuclei, you can begin to appreciate why physicists became alarmed about nuclear fission. If a small fraction of an Avogadro's Number of Uranium nuclei could be induced to undergo fission nearly simultaneously, then the release of energy (small for a single nucleus) would be multiplied by that huge number and a "super bomb" would result.

3) How much oil is consumed in the US? This is a timely and relevant question since the "Gulf crisis" and subsequent war six years ago.

Solution: Where do you begin such an estimate? Well, the use of oil (or its byproducts) with which you are probably most familiar is the use of gasoline in your car or your family's car. Approximating your gasoline consumption will be a good starting point. Then estimating the number of cars in the country will give the next bit of necessary data. A factor for other motor vehicles will be needed also, i.e. trucks, buses, airplanes. Finally a guess at the amount of gasoline produced by each gallon or barrel of crude oil will be needed. What about heating oil, oil for generating electricity, and other petroleum products like plastics?

To begin the oil consumption estimation consider a single personal automobile. If you own one (or have noticed what family or friends spend on gasoline), then you know that a typical fuel tank holds on the order of 10 gallons (and gas costs over \$1.20 per gallon as of Jun.11,1997). Furthermore such cars run about 200 or so miles on 1 tank, so that the "mileage" or fuel efficiency is about 20 miles per gallon (20 mpg). How many miles does a typical car cover in one year? There is obviously a lot of variation, but a typical distance (that the I.R.S. assumes average for auto expenses on the income tax forms) is 10,000 miles. Hence for 1 car the consumption is

10,000 miles
________________ = 500 gal in 1 year.
20 miles/gal

Next we need an estimate of how many cars are in use in the US. The population is about 250 million and perhaps 1 in 3 drive a car; say 100 million cars for simplicity. Then all passenger cars consume 500 x 100x106 = 5x104 x 106 = 5x1010 gallons of gasoline in 1 year. Now a barrel of oil is defined as 42 gallons, so assuming that most of the crude oil becomes gasoline we obtain

5x1010 gal/yr 50x109
___________ = ______ = 1.2x109 bbl/yr ,
42 gal/bbl 42

or in terms of barrels per day we have

1.2x109 bbl/yr 12x108
____________ @ ______ = 3x106 bbl/day.
365 day/yr 4x102

So 3 million barrels per day is our estimate, without taking account of trucks, buses, airplanes, heating oil and electricity generation. We also have not allowed for the fact that not all of the barrel of crude oil becomes gasoline, perhaps 2/3. Also a factor of 2 is probably appropriate for other motor vehicles (which are less numerous but travel further and are much less efficient) and another factor of 2 for heating and electricity. Multiplying all of these additional factors, 3/2 x 2 x 2 = 6, times the 3 million gives 18 million barrels per day.

Is that reasonable? Looking in an Almanac for 1995, which has petroleum consumption figures for 1993, we see that some 17 million barrels per day is the total figure. Not bad at all, considering how rough the approximations had to be!

To connect all of this to recent history, note that about 45% of that consumption is supplied by imports. OPEC countries produced about 18 million barrels per day, so the US must have bought about half of all the OPEC oil. With Iraq and Kuwait oil being roughly 20% of OPEC's production before the Gulf War, that means that about 10% of the US's normal supply was stopped with the embargo on Iraq(as of Sept.'90). How did that 10% affect the consumption? We know that the price of gasoline went up by some 40% shortly thereafter (but settled down again within a few months after the war), but did consumption decrease with the reduced supply? Or did oil production increase? How did this jibe with comments in the media about oil production at the time?