Philosophy 2801F/G Lecture Notes - Bridle

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18 Sep 2012

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The right of nature, the preservation of his own nature; that is to say of his own life. A law of nature general rule, found out by reason by which a man is forbidden to do, that which is destructive of his life, or taketh away that means of preserving the same. Every man has a right to everything even to any man. That every man ought to endeavour peace. Fundamental law of nature seek peace and follow it. Second nature by all means we can to defend ourselves. Whatsoever you require that others should do to you, that do ye to them. There is nothing to which every man had not right to nature. Right is laid aside, either by simply renouncing it or by transferring it to another. By simply renouncing when he cares not to whom the benefit thereof redoundeth. By transferring when he intendeth the benefit thereof to some certain person or persons.

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Background information
A person with normal vision can focus on objects as close as a fewcentimeters from the eye up to objects infinitely far away. Thereexist, however, certain conditions under which the range of visionis not so extended. For example, a nearsighted person cannot focuson objects farther than a certain point (the far point), while afarsighted person cannot focus on objects closer than a certainpoint (the near point). Note that even though the presence of anear point is common to everyone, a farsighted person has a nearpoint that is much farther from the eye than the near point of aperson with normal vision.

Both nearsightedness and farsightedness can be corrected with theuse of glasses or contact lenses. In this case, the eye convergesthe light coming from the image formed by the corrective lensrather than from the object itself.

Part A
When glasses (or contact lenses) are used to correctnearsightedness, where should the corrective lens form an image ofan object located at infinity in order for the eye to form a clearimage of that object?

The lens should form the image at the far point.

Correct
PArt B If a nearsighted person has a far point d_f thatis 3.50 m from the eye, what is the focal length f₁ thecontact lenses that the person would need to see an object atinfinity clearly?

f₁ = ? m

A nuclear reaction is represented by an equation similar to theequation for a chemical reaction. An example of a nuclear reactionis the decay of radium into radon by the emission of an alphaparticle (a helium nucleus). The equation for this reactionis

^{226}_{\ 88}{\rm Ra}\; \rightarrow \, ^{222}_{\ 86}{\rmRn}+\;^4_2{\rm He}.

The superscript preceding each element indicates the nucleon numberof that particular nucleus (total number of protons and neutrons).The subscript indicates the charge of the nucleus, which is equalto the atomic number for nuclei. Nuclear reactions frequentlyinvolve electrons (sometimes called beta rays) and photons(sometimes called gamma rays). Electrons, which have nucleon number0 and charge -1, are represented in equations as either ^{\ \;0}_{-1}\beta or ^{\ \; 0}_{-1}{\rm e}. Photons are represented as^{0}_{0}\gamma to emphasize that they have no nucleon number andcarry no charge.

In nuclear reactions, several conservation laws are usuallysufficient to determine the products or reactants. One of these isconservation of nucleon number. The total number of neutrons andprotons in all of the reactants must be the same as the totalnumber of protons and neutrons in the products. Although protonsand neutrons sometimes change into one another, neither can simplydisappear. Another important conservation law is conservation ofcharge. The net charge of the reactants must be the same as that ofthe products.

Notice that, in our equation for the decay of radium, the sum ofthe nucleon numbers on each side is 226 and the sum of the chargeson each side is 88. If you knew everything in this equation exceptthe isotope of radon produced, you could simply subtract 4 from 226to find the nucleon number required to make the nucleon number ofthe products equal the nucleon number of the reactants. Similarly,if you knew everything in this equation except for the identity ofthe product element, you could simply subtract 2 from 88 and thenlook in a periodic table to find that the element with atomicnumber 86 is radon.

In parts A and B consider the decay of radon by alpha particleemission:

^{222}_{\ 86}{\rm Rn}\;\rightarrow\;X +\,^{4}_{2}{\rm He}.

a. What is the nucleon number of the element X?
Express your answer as an integer.

b. What is the atomic number (nuclear charge) of the elementX?
Express your answer as an integer.

n parts C and D consider the decay of Potassium-43 into calcium-43via the following reaction:

^{43}_{19}{\rm K} \rightarrow ^{43}_{20}{\rm Ca}+\,X.

c. What is the charge of the emitted particle X?
d. is not visible yet

E. An isotope of potassium with a half-life of roughly one billionyears is often used in radioactive dating. This isotope decaysthrough electron capture, the process of drawing an electron intothe nucleus. What is the chemical symbol for the element thatresults from this process? The equation for the reaction is

^{40}_{19}{\rm K}+\,^{\ \;0}_{-1}{\rm e}\rightarrow \,?.
Express your answer as a chemical symbol (e.g., Na forsodium).

f. The reaction energy of a reaction is the amount of energyreleased by the reaction. It is found by determining the differencein mass between the reactants and products and then using E=mc^2 toconvert the lost mass into the released energy.

Look again at the equation for the decay of radium (Ra), given inthe introduction. Use the following table of masses to determinethe reaction energy of this reaction.
Nucleus or particle Mass in atomic mass units
radium-226 226.025402
radon-222 222.017571
alpha particle 4.002602
Express your answer in joules to three significant figures.

please answer fully... thank you

3. 3 Stratosphere-troposphere exchange

The rate of exchange of air between the troposphere and thestratosphere is critical for determining the potential of variouspollutants emitted from the surface to reach the stratosphere andaffect the stratospheric ozone layer. One of the first estimates ofthis rate was made in the 1960s using measurements of strontium-90(90Sr) in the stratosphere. Strontium-90 is a radioactive isotope(half-life 28 years) produced in nuclear explosions. It has nonatural sources. Large amounts of 90Sr were injected into thestratosphere in the 1950s by above-ground nuclear tests. Thesetests were banned by international treaty in 1962. Following thetest ban the stratospheric concentrations of 90Sr began to decreaseas 90Sr was transferred to the troposphere. In the troposphere,90Sr is removed by wet deposition with a lifetime of 10 days (bycontrast there is no rain, and hence no wet deposition, in thestratosphere). An intensive stratospheric measurement network wasoperated in the 1960s to monitor the decay of 90Sr in thestratosphere. We interpret these observations here using a 2-boxmodel for stratosphere-troposphere exchange with transfer rateconstants kTS and kST (yr-1) between the tropospheric andstratospheric reservoirs. The reservoirs are assumed to beindividually well-mixed.

Let mS and mT represent the masses of 90Sr in the stratosphere andin the troposphere respectively. Observations of the decrease inthe stratospheric inventory for the period 1963-1967 can be fittedto an exponential mS(t) = mS(0)exp(-kt) where k = 0.77 yr-1.

1. Write mass balance equations for mS and mT in the 1963-1967period.

2. Assuming that transfer of 90Sr from the troposphere to thestratosphere is negligible (we will verify this assumption later),show that the residence time of air in the stratosphere is tS =1/kST = 1.3 years.

3. Let m'T and m'S represent the total masses of air in thetroposphere and the stratosphere, respectively. Show that theresidence time of air in the troposphere is tT = tS (m'T/m'S) = 7.4years. Conclude as to the validity of your assumption in question2.

4. Hydrochlorofluorocarbons (HCFCs) have been adopted asreplacement products for the chlorofluorocarbons (CFCs), which werebanned by the Montreal protocol because of their harmful effect onthe ozone layer. In contrast to the CFCs, the HCFCs can be oxidizedin the troposphere, and the oxidation products washed out byprecipitation, so that most of the HCFCs do not penetrate into thestratosphere to destroy ozone. Two common HCFCs have trade namesHCFC-123 and HCFC-124; their lifetimes against oxidation in thetroposphere are 1.4 years and 5.9 years, respectively. There are noother sinks for these species in the troposphere. Using our 2-boxmodel, determine what fractions of the emitted HCFC-123 andHCFC-124 penetrate the stratosphere.

Philosophy 2801F/G Lecture Notes - Bridle

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