Section   10.1The Atomic Nucleus

1. (Exercises 3 and 4) If 51.83% of all silver found on Earth is ¹⁰⁷Ag (atomic mass = 106.905 u) and the other 48.17% is ¹⁰⁹Ag (atomic mass = 108.905 u), what is the atomic mass of silver?

Using the fractional abundances given above, we can find the atomic mass of silver as follows:

0.5183 x 106.905 u	=	55.41 u	for 107Ag
0.4817 x 108.905 u	=	52.46 u	for 109Ag
		107.87 u

Adding these together we get 107.87 u for the atomic mass of naturally occurring silver.

Section   10.2Radioactivity and Half-life

2. (Exercises 7 and 8) Indium-115 is radioactive and undergoes beta decay. Write the equation for this reaction.

Indium has 66 neutrons and 49 protons in its nucleus. During beta decay, one of these neutrons is transformed into a proton. This means that the mass number remains the same, but the atomic number increases by one. In this process an electron is emitted. The equation can then be written as follows:

¹¹⁵₄₉ In --> ¹¹⁵₅₀Sn + ⁰_-1e

Note that the daughter nucleus is now another element, tin, ¹¹⁵₅₀Sn, and that the emitted particle is an electron, ⁰_-1e. For Exercise 7 a. in the textbook, the reaction is alpha decay and the daughter nucleus is two places lower in the periodic table than its parent because it has two fewer protons after decay. The mass number decreases by four because two neutrons and two protons (four nucleons) have been lost. The decay particle in this case is composed of two protons and two neutrons and is called an alpha particle. An alpha particle is identical to the nucleus of a helium atom (two protons + two neutrons). The decay equation is therefore:

²²⁶₈₈Ra --> ²²²₈₆Rn + ⁴₂He

3. (Exercises 11 and 12) Chlorine-34 is a radionuclide with a half-life of 33 minutes.

a. How many half-lives will it take for the count rate on a specific sample of this radionuclide to decrease from 4000 cpm to 62.5 cpm?

b. How long will it take for the decay rate to decrease to 62.5 cpm?

a. During each half-life, the decay rate will decrease by a factor of one-half, so the following calculation shows all decay periods during this reduction in decay rate.

1/2 4000 cpm = 2000 cpm		Note that each line represents a decrease in count rate by a factor of one-half, so each line indicates the passage of one half-life. Since there are 6 lines in which the count rate decreases by one half, a total of 6 half-lives have passed.
1/2 2000 cpm = 1000 cpm
1/2 1000 cpm = 500 cpm
1/2 500 cpm = 250 cpm
1/2 250 cpm = 125 cpm
1/2 125 cpm = 62.5 cpm

b. The length of one half-life is given as 33 minutes, so six half lives will take 6 half-lives x 33 minutes/half-life, or 198 minutes.

Section   10.4Nuclear Reactions

4. (Exercises 21 and 22) Complete the following nuclear reaction equations:

a. ¹²₆C + ¹₁H --> ______ + gamma ray

b. ²₁H + ¹⁴₇N --> ¹²₆C + ______

c. ⁴₂He + ⁹₄Be --> ¹²₆C + ______

In each case the number of nucleons (the sum of the superscripts) on one side of the reaction arrow must equal the number of nucleons on the other side, and the total atomic number (the sum of the subscripts) on one side of the reaction arrow must equal the total atomic number on the other side.

For case a, 12 + 1 on the left side means that the superscript on the unknown atom on the right must be 13, and 6 + 1 on the left side means that the atomic number of the atom on the right side must be 7. This makes the unknown atom ¹³₇N, because the gamma ray has 0 for its mass number and 0 for its atomic number. The full equation is then:

a. ¹²₆C + ¹₁H --> ¹³₇N + gamma ray

For case b, 2 + 14 = 16 on the left side means that the superscript on the unknown atom on the right must be 16 - 12 = 4, and 1 + 7 = 8 on the left side means that the atomic number of the atom on the right side must be 8 - 6 = 2. This makes the unknown atom ⁴₂He. The full equation is then:

b. ²₁H + ¹⁴₇N --> ¹²₆C + ⁴₂He

For case c, 4 + 9 = 13 on the left side means that the superscript on the unknown particle on the right must be 13 - 12 = 1, and 2 + 4 = 6 on the left side means that the atomic number of the particle on the right side must be 6 - 6 = 0. This makes the unknown particle ¹₀n. The full equation is then:

c. ⁴₂He + ⁹₄Be --> ¹²₆C + ¹₀n

Section   10.5Nuclear Fission

5. (Exercises 23 and 24) Complete the following equation for nuclear fission.

¹₀n + ²³⁵₉₂U --> ²³⁶₉₂U --> ¹⁴⁰₅₄Xe + _____ + 2 ¹₀n

Using the same reasoning as we used in exercise 4, the superscript and subscript for the unknown atom must be 94 and 38, respectively. Atomic number 38 means that the atom must be strontium, and the full isotope designation must be ⁹⁴₃₈Sr, so the complete equation is:

¹₀n + ²³⁵₉₂U --> ²³⁶₉₂U --> ¹⁴⁰₅₄Xe + ⁹⁴₃₈Sr + 2 ¹₀n

Section   10.6Nuclear Fusion

6. (Exercises 25 and 26) In older stars, the fusion of heavy nuclei with alpha particles continues to build up heavier elements in the periodic table. When an alpha particle (a helium-4 nucleus) combines with a nitrogen-14 nucleus in a particle accelerator, the result is the formation of an oxygen-17 nucleus, and in the process a proton (hydrogen-1 nucleus) is also produced. Given the following atomic masses,

a. calculate the mass difference in u,

b. determine the energy produced in MeV, and

c. state whether the interaction is endoergic or exoergic.

The reaction under study here is the first one considered in exercise 19 in the textbook, and it can be written as follows. The atomic mass of each nucleus is written below its symbol.

42He	+	147N	-->	178O	+	11H
(4.00260 u)		(14.00307 u)		(16.99916 u)		(1.00783 u)

a. In order to find the mass difference we must add all of the atomic masses on each side of the arrow and subtract the right-hand total from the left-hand total.

18.00567 u - 18.00699 u = - 0.00132 u

This indicates that the products of this reaction are slightly heavier than the sum of the component parts from which they are formed. This means that mass, in the form of energy, must be supplied by a particle accelerator so that the reaction can take place. b. The amount of energy that must be supplied can be found by using Einstein's mass/energy relationship, E = mc², where we know from the discussion in the textbook that each u is equivalent to 931 MeV.

0.00132 u x 931 MeV/u = 1.23 MeV of energy must be supplied

c. As stated above, energy must be added to make the reaction occur, so this nuclear interaction will be endoergic.

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