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An Introduction to Physical Science, Ninth Edition
James T. Shipman, Ohio University
Jerry D. Wilson, Lander University
Aaron W. Todd, Middle Tennessee State University
Solved Problems

Chapter 10: Nuclear Physics

Neutral atoms are made up of electrons, protons, and neutrons. Atomic notation tells us exactly how many of each at a glance.

1. Given the neutral iron atom 5626Fe, indicate the number of protons, neutrons, and total nucleons in the nucleus, and the number of electrons in orbit around the nucleus.

From the designation we can see that the atomic number to the left of the chemical symbol is 26, so there are 26 protons in the nucleus. The mass number is 56, so the total number of nucleons is 56. The number of neutrons is the mass number minus the atomic number, that is 56 - 26, so there are 30 neutrons in the nucleus. For a neutral atom, the number of electrons in orbit must be exactly equal to the number of protons present in the nucleus, so the number of electrons in this neutral atom is also 26. If the iron had been ionized it would have gained or lost one or more electrons. Singly ionized iron would have only 25 electrons, but the number of protons would remain the same. If the number of neutrons was different, we would have another isotope of iron.


2. Write the atomic notation for an atom of iron that has 28 neutrons in its nucleus.

Since we still have an atom of iron, the atomic number must still be 26. The new mass number will be 26 + 28 = 54. The correct symbolic notation will then be:

5426Fe
Now let's look at the stability of an atomic nucleus. The textbook gives some general criteria for nuclear stability that we can apply to determine if a particular nuclide will be stable or unstable.


3. Why do you think the three nuclides listed below are radioactive?
a. 22686RnNuclides with atomic numbers greater than 83 have so many positively charged protons in their nucleus that they are always radioactive.
b. 116CThe number of neutrons is fewer than the number of protons. The only nuclides that are stable with fewer neutrons than protons in their nucleus are hydrogen-1 and helium-3.
c. 5625MnThis isotope has an odd number of protons (25) and also an odd number of neutrons (31). Odd-Odd nuclei are generally unstable.

4. A fossil is the remains of a once-living organism that contained nearly a fixed ratio of 12C to radioactive 14C while it was alive. If we now measure the beta emission from the fossil remains of this organism to be 1/8 of the rate found in living organisms, about how old is the fossil?

Because the half life of carbon-14 is known to be 5730 years, we can find out how long it has been since the fossil was alive if we know the number of half-lives that has passed. In one half-life the count rate would have decreased to 1/2, in the second half-life from 1/2 to 1/4, in the third half-life from 1/4 to 1/8. The fossil has been dead for three half-lives of the carbon-14, or:

time = 3 half-lives x 5730 years/half-life = 17190 years
The fossil has, therefore, probably been dead for a little over 17 thousand years.

In nuclear decay equations, the sums of mass numbers and the sums of atomic numbers on each side of the reaction arrow must be the same. This means that we can predict what particle will be given off in any specific case if we know the parent and daughter nuclei involved in the reaction. Exercise 6 in the textbook gives you several reactions to analyze. Let's take a look at number 6(b).


5. Complete the following nuclear decay schemes, and state whether the process is alpha decay, beta decay, gamma decay, or fission.

21084Po --> 20682Pb + _____
The total mass number on the left minus the known mass numbers on the right must equal the mass number of the missing particle. In this case, 210 - 206 = 4. The atomic number can be determined in a similar way; 84 - 82 = 2.

The decay particle must have a mass number of 4 and an atomic number of 2. This means the particle is made up of 2 protons and 2 neutrons, which is a stable decay product called an alpha particle. The process is, therefore, alpha decay.

21084Po --> 20682Pb + 42He
Now let's try another example.
24094Pu --> 9738Sr + 13956Ba + _____
The total mass number on the left minus the known mass numbers on the right must equal the mass number of the missing particle(s). In this case, 240 - (97 + 139) = 4. The atomic number can be determined in a similar way: 94 - (38 + 56) = 0.

At first, this appeares to be another alpha decay, because the mass number difference comes out to be 4, but since the atomic number difference is 0 there can be no protons (charge +1) in the products, so we must have only neutrons. For this reaction there must be 4 separate neutrons produced, each of one mass unit, and the reaction is a fission decay process.

24094Pu --> 9738Sr + 13956Ba + 410n


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