Chapter 31
The Giancoli onLine Tutor

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Nuclear Energy; Effects and Uses of Radiation.

I. Nuclear Reactions

A Nuclear Reaction occurs when a nuclear particle (usually a neutron, proton or alpha particle) interacts with a Nucleus. The result (usually) will be another nucleus and another nuclear particle. Charge is conserved so the number of protons must be the same on both sides of the reaction, also, the total number of protons and neutrons should be the same on both sides. Note to preserve the number of electrons, use ¹₁H instead of p and ⁴₂He instead of a.

Interactive Activity
Choose each reaction in turn below and fill in the empty blocks with the correct entries. When you think you have them correct click the grade button and any mistakes will be erased giving you a chance to correct your entries.

+ -> +

The Q-value of a reaction is the difference in mass energy before and after. If the reaction is
x + X -> Y + y
then
Q = (M_x +M_X - M_Y-M_y)c²
Calculate the Q for each of the four reactions above and check your results below. (Be sure to use atomic masses, e.g., ¹₁H not p, and use the value in atomic units . Use the data given in the Appendix. Click the final field for your result.

M_x =

M_X =

M_Y =

M_y =

DM = xc² = Q =

If the Q-value is negative the reaction is endothermic (meaning that energy has to be added to have the reaction go. For example, if Q = -3 MeV, at least that much energy must be supplied ( by the incident particle x ). Actually more than that would be needed in order to conserve momentum. The threshold energy, K_th, would be
K_th = -Q(1 + M_x/M_X)

A particular type of Nuclear reaction is a fission reaction when a neutron hits a heavy nucleus such as ²³⁵U and causes it to split into two smaller nuclei (generally one 'bigish' and one 'smallish') along with 2 or 3 neutrons. The extra neutrons can cause cascading fissions (a chain reaction) in nearby nuclei.

Interactive Activity
Below is a model of this, each checkbox is a 'nucleus'. Click any one of them to initiate the 'Chain Reaction'.

REDO

You probably noticed that after the first 80% of the fissions things really slow down (because the targets become few). In real life not all the fissile material actually undergoes fission, however, in the demo, there is a 'surprise' for you if you wait to the end.
There are a lot of factors that determine the success of a chain reaction, one important factor is the critical mass. There is a minimum size necessary for the chain reaction to occur before the neutrons escape the mass.

II. Nuclear Power

In a Nuclear Power Plant a small amount of ²³⁵U is included with a large amount of non-fissioning ²³⁸U. Water under high pressure is used to remove the heat energy. It also acts as a moderator to slow the neutrons down so that they don't interact with the ²³⁸U. A schematic model follows:

To find out more about how this works go here and there.

A power plant like the one above is about 33% efficient, so that to produce 1000 MW (1x10⁹ J/s )of electrical power it will need 3000 MW of nuclear energy. A typical fission reaction is
n + ²³⁵U -> ¹⁴¹Ba + ⁹²Kr + n + n + n
The mass of the daughter products is usually not known (they don't exist long enough) . However, nucleons (protons and/or neutrons) in the midrange nucei are bound tighter by about .9 MeV, which means that each fission releases about 200 MeV. A pound (.45 kg) of U contains 0.7 % ²³⁵U. which means 1 pound contains about 1x10²⁴ ²³⁸U atoms and 8x10²¹ ²³⁵U atoms. How did you do that?.

If all the ²³⁵U atoms fission in one second that would produce 8x10²¹ fissions/s*200 MeV =
1.6x10²⁴ MeV/s(*1.6x10^-13 J/MeV) =
2.56x10¹¹ J/s=
256000x10⁶ J/s =
256000 MW
or almost 100 times what is need to furnish 1000 MW electric.

Putting it another way, one pound of U would run the power plant for a little over a minute and a half.

The energy-to-mass ratio for U is 2.56x10¹¹ J/0.45 kg = 5.7x10¹¹ J/kg. By contrast that ratio for coal is about 2.9x10⁷ J/kg and for wood 1.8x10⁷ J/kg.

III. Nuclear Fusion

As impressive as the energy-to-mass ratio is for Uranium fission fuel, it is even more so for fusion. Deuterium (²₁H) is comonly available in seawater and the reaction

²₁H + ²₁H -> ³₂He + n
produces 3.27 Mev ( 5.23x10^-13 J) and the fuel has a mass of 4 u (6.67x10^-27 kg). This is a mass-to energy ratio of 7.9x10¹³ J/kg.

The fission reaction occurs when an uncharged neutron enters a charged nucleus, the fusion reaction occurs when two charged nuclei come in contact. It takes about 50 pounds of force to bring two protons together . This force is supplied by heating a gas until all electrons are stripped from the atoms thus creating a plasma. The plasma is then heated until the random kinetic energy of motion ( 3kT/2, k= Boltzmann's Constant=1.38x10^-23 J/K) overcomes the repulsive electric potential energy (ke²/r, k= Coulombs Constant=9x109 Nm²/C²). Setting twice the KE (both particles have kinetic energy) equal to the potential energy we obtain
T = 5.56x10^-6/(r₁+r₂),
where r₁, r₂ are the radii of the particles involved. They are calculated by using r = 1.2x10^-15*A^1/3.

Interactive Activity
Calculate the needed temperature for various combinations of ¹₂H, ²₁H, and ³₁H, check your answer below.

r₁+r₂ = x10^-15 m

T = K

IV. Radiation

Radiation Units:

UNIT	SYMBOL	USE	DEFINITION	Type	COMMENT
curie	Ci	measure source strength	3.70x10¹⁰ disintegrations/s	Traditional Unit	In very common use.
becquerel	Bq	measure source strength	1 disintegration/s	SI unit	not common
roentgen	R	Absorbed Dose	0.878 x10^-2J/kg air	Out of Date unit	The number refers to energy created by radiation ionizing the air as it passes through it. (For X-rays and g rays.)
rad	rad	Absorbed Dose	0.01 J/kg material	Traditional unit	In common use
gray	Gy	Absorbed Dose	1 J/kg materisl	SI unit	Becoming more common
rem	rem	Effective Dose	rad x QF	Traditional unit	Some radiation does more damage than others. This is measured by the Quality Factor. The QF for X-rays is 1 and that for alpha radiation about 20
sievert	Sv	Effective Dose	Gy x QF	SI unit	not common

Radiation Dose

Calculating raddiation dose can be tricky. It depends upon:

The Source Strength (Ci)
- The source should be converted to disintegrations/s
Radiation Energy
- Multiply the source strength by the energy of the radiation.
  
  For example a 1 Ci source of 1 Mev g rays would be emitting 3x10¹⁰*1 MeV*1.6x10-13 J/Mev= 4.8x10^-3 J/s
Insulation (Shielding)
- Not covered at this level
Air Absorbtion
- An inch or so of air will stop a radiation. A few feet will do the same for b radiation. Air can be ignored for X and g

Distance from source

If the target has a surface area of A m², then the amount of radiaion energy intercepted at a distance R will be A/4pR². For example if A=1 and R=3, using the previous numbers, we obtain 4.8x10^-3 J/s*1/4p3² = 4.24x10^-5 J/s

Penetrability of the radiation

For a and b radiation all would be absorbed. (If they even reached the body. They are mainly a danger when the source is ingested.) For g and X-rays assume 50%. For the above numbers then, 4.24x10^-5 J/s*.5(50%)=2.12x10^-5 J/s deposited in the body.

Body mass

For g and X-rays you can assume the radiation is distributed througout the body. Thus using the above numbers, for a 50 kg body the absorbed dose rate is 2.12x10^-5 J/s /50 kg = 4.24x10-7 J/kgs = 4.24x10-7 Gy/s

Time
Multiply the previous result by the exposure time in seconds.

For a 24 hour exposure to the above, 4.24x10-7x24x60x60 = .037 Gy= 3.7 rad.

Interactive Example
A laboratory worker is inadvertantly exposed to an unshielded 5 Ci source emitting 2 MeV gamma rays for 10 hours over the course of a week. Calculate the dose in rads by following the steps below.

What is the power (J/s , to four decimal places) emitted by the source in gamma ray radiation. Enter your answer and click below.

If the worker has a body area of 2 m² and is 10 meters from the source, what fraction of the emitted power (to the nearest tenth of a microwatt) reaches him? Enter your answer below and click outside.

Assuming 100 kg body weight and 50% penetrability what is the absorbed dose per second to the nearest nanoGy per second? Enter your answer below and click outside.

What is the total exposure to the nearest milliGy ? Enter your answer below and click outside.

What is the total exposure to the nearest tenth of a rad ? Enter your answer below and click outside.