Chapter 30
The Giancoli onLine Tutor
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Nuclear Physics and Radioactivity.

I. Nuclear Nomenclature

Any nucleus is uniquely determined by two numbers.
Z is known as the atomic number, A is called the Atomic Mass Number, the chemical symbol, X ( eg, Cl, U) , is in a sense redundant since it is uniquely determined by Z.

Interactive Examples

Two common isotopes of Uranium are 23592U and 23892U:

How many protons does Uranium have (Click feedback below to find out your result after each question)?

How many neutrons does 235U have?

How many neutrons does 238U have?

II. Nuclear Size

The nucleons (protons-red, neutrons-blue) in a nucleus are closely packed, consequently the size,volume, is proportional to the number of particles, A, and, assuming a sprerical like shape (V=4pr3/3) the radius will be proportional to A1/3. It is found that
r = 1.2x10-15*A1/3.

Interactive Examples

What is the Chemical Symbol for the nucleus to the right above (Click feedback below to find out your result after each question)?

What is its presumed radius in fm to 2 decimal places?

If the nucleus does not have the correct balance of protons to neutrons it will be radioactive. Is the pictured nucleus radioactive (yes or no)?

III. Nuclear Mass and Binding Energy

The mass of the two "stuck-together" horseshoe magnets at the left is actually less than the sum of the masses of each (though you will never prove it by any measurement you might make). The mass of the combination is reduced by the binding energy divided by c2. For magnets this is a very small number indeed, but for nucleons held together by the very strong Nuclear Force the story is much different.

Nuclear and atomic masses are commonly given in atomic mass units (u), 1 u= 1.6605x10-27 kg. The atomic mass unit is chosen so that the mass of 12C would be exactly 12 (see below).
Element Z A Mass (u)
C 6 12 12.000000
C 6 14 14.003242
N 7 14 14.003074


To find the binding energy of calculate the difference in the mass of the nucleus M(Z,A) and the mass of its constituents.
BE/c2 = (A-Z)M(n) + ZM(11H) - M(Z,A)
use c2 = 931.5 MeV/u.
M(n) is the mass of the neutron (1.008665 u); M(11H) is the mass of the hydrogen atom (1.007825). The Mass of the hydrogen atom is used instead of the mass of the proton because the mass of the nucleus is actually the mass of the neutral atom including its Z electrons. By using the mass of Z hydrogen atoms we subtract out Z electron masses from M(Z,A).

Interactive Example

Calculate the binding energy for 14C and 12C and enter your results below.
Binding Energy in MeV(1 decimal place) for 14C (Click feedback below to find out your result after each question)?


Binding Energy in MeV(1 decimal place) for 12C?


What is the binding energy per nucleon for 12C?


Semiempirical Mass Formula

There is a rather complicated formula based on a model of the nucleus that treats it as a very dense liquid drop. This model is not very good for light nuclei but, excepting these, it does a pretty good job over a wide range of nuclei. You can explore its predictions below.

Interactive Activity

You can calculate (approximately) the binding energy/nucleon by entering Z and A below and clicking in the result field. See if you can find the nucleus with the highest value for this number (this will be the most stable nucleus). When you think you know the answer enter it (Z,A), without the (), in the answer field and click outside. Hint: For lighter nuclei Z~(A-Z) and for very heavy nuclei Z~(A-Z)*2/3. Note: The results of the calculation are low for light nuclei like 12C (check it) but get more accurate for heavier nuclei, for example, the correct value for 238U is 7.57 MeV.

Z= A=
Result (BE/A)=
Answer=

IV. Nuclear Stability and Radioactivity

Beta Decay

If the nucleus does not have the correct ratio of neutrons to protons , it will readjust the ratio using radioactivity.
The picture shows a nucleus (14C) changing into (14N) by a neutron changing into a proton and emitting an electron in the process. The electron is called a beta particle in this context, and the process is called beta decay. If you look closely you will see that there is another particle also emitted, n, called the neutrino (actually because of the bar over it is technically an antineutrino). Other than explaining why the electron does not carry off all the energy released in the decay, the neutrino is not important to this chapter.

Interactive Example

What is the energy released in 14C decay (disintegration energy aka Q-value) to the nearest kev? The mass of the electron is 0.000549 u and the mass and charge of the neutrino are zero. (Hint: Use E = Dmc2 and remember that the given masses are for the neutral atom with all its electrons. N has 1 more electron than C. )




Alpa Decay

If you don't have the Quicktime plugin there is probably a broken image at the left. You can view a GIF animation version of the movie here.
A heavy nucleus has many more neutrons than protons, if the ratio of protons to neutrons is too rich it can be reduced by the emission of an a particle ( two neutrons and two protons,i.e., the very stable nucleus of the helium atom.

Say that there is a nucleus with 4 protons and 6 neutrons, that is a proton to neutron ratio of 2/3. If two protons and two neutrons are removed the new ratio is decreased to 1/2. That is an example of why in a heavy nucleus where the number of neutrons exceeds the number of protons loosing two of each decreases the ratio of protons to neutrons.

As shown in the movie, two protons and two neutrons tend to cluster together in heavy nuclei and, if possible, will eventually 'break-out' creating a process called alpha decay. Most often this leaves the daughter nucleus in an excited and the alpha emission is followed by a gamma ray ( high frequency electromagnetic photon). Unlike beta decay in which Z increase by 1, in alpha decay A decreases by 4 and Z by 2.

Interactive Example

235U (235.043925 u ) undergoes alpha (4.002602 u ) decay: What is the symbol for the resulting element? (See appendix)


What is the mass number of the resulting element?


How much energy to the nearest MeV is released in the decay?




Positron Decay and Electron Capture

If the nucleus is light, and the number of protons and neutrons are approximately equal, other means are available for decreasing the proton to neutron ratio. For 40K (39.963999) accomplishes this via positron decay ( the positively charged antielectron). Inside the nucleus a proton changes into a neutron and a positron is emitted (also a neutrino). Alternately, electron capture can, also, occur. Electrons in a s subshell (zero angular momentum) actually pass through the nucleus and can be captured by a proton changing it into a neutron (and emitting a neutrino). In both cases A stays the same and Z decreases by 1.

Interactive Example

In 40K decay what is the symbol for the resulting element? (See appendix)

What is the energy (in MeV) released in the positron decay? Hint: since Z is decreased by one, there will be an extra atomic electron left over as well as the positron.

What is the energy (in MeV) released in the electron capture? Hint: a nucleus with Z electrons captures one of its atom's electrons and changes into a nucleus with one less electron in its atom.



V. Radioactivity

The formula on the left is the radioactive decay formula. It applies to any collection of objects that decrease in a fixed fraction. For example if you shake a box of 100 coins once every second and remove all those that come up heads, you would expect to have 50 after 1 shake, 25 after the second, 12 or 13 after the third and so on. On the other hand if it was 100, 6 sided dice and remove all that come up 1 you would expect to have left 83 then 69 and so on. Of course with small numbers there would be fluctuations about this but the trend would be the same. The given equation should mirror this result, however, it is developed using the premise that there is a huge number to start with. Below is an example in which you can control the starting number and the decay constant (which is no more than the probability of an event...1/2 for coins. 1/6 for regular dice, etc.).

Interactive Activity

The column on the left just uses the probabilities as we did above. The column on the right uses the formula. As you can see for large N and small l they very closely parallel! (Actually the results are not bad for small N and small l).
Decay Probability: Starting number:


Radioactive Half-Life

The half-life, T1/2 is given by

T1/2 = .693/l
Thus after 7 'shakes' with l = 1/10 (T1/2=6.93) you should have about 1/2 of what you started with, after another half-life you should have 1/4 of your starting value and so on. CHECK THIS OUT! What should be the half life for the other values of l, check your calculation by running the simulation above.

The actual radioactivity (usually said as activity) at any time is lN.

Interactive Example

What is the activity at 4 seconds, (using the right hand column above) with l = 1/10 (units = sec-1, so your answer will be in #/sec usually expressed as decays/sec) and No = 1000000?

The half-life of 131I is 8.05 days, what is the value for l in sec-1?.

If you have 2 mg of 131I how many atoms do you have? (Hint: What is Avogadro's number?)

What is the activity of your I sample ?


PART II


© 1999 Carl Adler mailto:Carl@Image-ination.com