Sample Problem 9.1 Radiation Protection

1. Sample Problem 9.1Radiation Protection How does the type of shielding for alpha radiation differ from that used for gamma radiation? Solution Alpha radiation is stopped by paper and clothing. However, lead or concrete is needed for protection from gamma radiation. Study Check 9.1 Besides shielding, what other methods help reduce exposure to radiation?

2. Sample Problem 9.2Writing an Equation for Alpha Decay Smoke detectors used in homes and apartments contain americium-241, which undergoes alpha decay. When alpha particles collide with air molecules, charged particles are produced that generate an electrical current. If smoke particles enter the detector, they interfere with the formation of charged particles in the air, and the electric current is interrupted. This causes the alarm to sound, and warns the occupants of the danger of fire. Complete the following nuclear equation for the decay of americium-241: Solution Step 1 Write the incomplete nuclear equation. Step 2 Determine the missing mass number. In the equation, the mass number of americium, 241, is equal to the sum of the mass numbers of the new nucleus and the alpha particle. Step 3 Determine the missing atomic number. The atomic number of americium, 95, must equal the sum of the atomic numbers of the new nucleus and the alpha particle. Step 4 Determine the symbol of the new nucleus. On the periodic table, the element that has atomic number 93 is neptunium, Np. The nucleus of this isotope of Np is written as .

3. Sample Problem 9.2Writing an Equation for Alpha Decay Continued Step 5 Complete the nuclear equation. Study Check 9.2 Write a balanced nuclear equation for the alpha decay of polonium Po-214.

4. Sample Problem 9.3Writing an Equation for Beta Decay Write the balanced nuclear equation for the beta decay of cobalt-60. Solution Step 1 Write the incomplete nuclear equation. Step 2 Determine the missing mass number. In the equation for beta decay, the mass number of cobalt, 60, is equal to the sum of the mass numbers of the new nucleus and the beta particle. Step 3 Determine the missing atomic number. The atomic number of cobalt, 27, must equal the sum of the atomic numbers of the new nucleus and the beta particle. Step 4 Determine the symbol of the new nucleus. On the periodic table, the element that has atomic number 28 is nickel (Ni). The nucleus of this isotope of Ni is written as

5. Sample Problem 9.3Writing an Equation for Beta Decay Continued Step 5 Complete the nuclear equation. Study Check 9.3 Write the balanced nuclear equation for the beta decay of chromium-51.

6. Sample Problem 9.4Writing an Equation for Positron Emission Write the balanced nuclear equation for manganese-49, which decays by emitting a positron. Solution Step 1 Write the incomplete nuclear equation. Step 2 Determine the missing mass number. In the equation, the mass number of manganese, 49, is equal to the sum of the mass numbers of the new nucleus and the positron. Step 3 Determine the missing atomic number. The atomic number of manganese, 25, must equal the sum of the atomic numbers of the new nucleus and the positron. Step 4 Determine the symbol of the new nucleus. On the periodic table, the element that has atomic number 24 is chromium, Cr. The nucleus of this isotope of Cr is written as

7. Sample Problem 9.4 Writing an Equation for Positron Emission Continued Step 5 Complete the nuclear equation. Study Check 9.4 Write the balanced nuclear equation for xenon-118, which decays by emitting a positron.

8. Continued Sample Problem 9.5Producing Radioactive Isotopes Step 5 Complete the nuclear equation. Study Check 9.5 The first radioactive isotope was produced in 1934 by the bombardment of aluminum-27 by an alpha particle to produce a radioactive isotope and one neutron. What is the balanced nuclear equation for this transmutation?

9. Sample Problem 9.5Producing Radioactive Isotopes Write the balanced nuclear equation for the bombardment of nickel-58 by a (11H) proton which produces a radioactive isotope and an alpha particle. Solution Step 1 Write the incomplete nuclear equation. Step 2 Determine the missing mass number. In the equation, the sum of the mass numbers of the proton, 1, and the nickel, 58, must be equal to the sum of the mass numbers of the new nucleus and the alpha particle. Step 3 Determine the missing atomic number. The sum of the atomic numbers of the proton, 1, and nickel, 28, must equal the sum of the atomic numbers of the new nucleus and the alpha particle, 2. Step 4 Determine the symbol of the new nucleus. On the periodic table, the element that has atomic number 27 is cobalt, Co. The nucleus of this isotope of Co is written as

10. Sample Problem 9.6Radiation Measurement One treatment for bone pain involves intravenous administration of the radioisotope phosphorus-32, which is incorporated into bone. A typical dose of 7 mCi can produce up to 450 rad in the bone. What is the difference between the units of mCi and rad? Solution The millicuries (mCi) indicate the activity of the P-32 in terms of the number of nuclei that break down in 1 second. The radiation absorbed dose (rad) is a measure of amount of radiation absorbed by the bone. Study Check 9.6 If phosphorus-32 is a beta emitter, how do the number of rems compare to the rads?

11. Sample Problem 9.7Using Half-Lives of a Radioisotope Phosphorus-32, a radioisotope used in the treatment of leukemia, has a half-life of 14.3 days. If a sample contains 8.0 mg of phosphorus-32, how many milligrams of phosphorus-32 remain after 42.9 days? Solution Step 1State the given and needed quantities. Given 8.0 mg of 3215P; 42.9 days; 14.3 days/half-life Need mg of 3215P remaining Step 2Write a plan to calculate the unknown quantity. Step 3Write the half-life equality and conversion factors.

12. Sample Problem 9.7Using Half-Lives of a Radioisotope Continued Step 4Set up the problem to calculate amount of active radioisotope. First, we determine the number of half-lives in the amount of time that has elapsed. Now, we can calculate how much of the sample decays in 3 half-lives and how many milligrams of the phosphorus remain. Study Check 9.7 Iron-59 has a half-life of 44 days. If the laboratory received a sample of 8.0μgof iron-59, how many micrograms of iron-59 remain active after 176 days?

13. Sample Problem 9.8Medical Application of Radioisotopes In the treatment of abdominal carcinoma, a person is treated with seeds of gold-198, which is a beta emitter. Write the balanced nuclear equation for the beta decay of gold-198. Solution We can write the incomplete nuclear equation starting with gold-198, which has the atomic number 79. In beta decay, the mass number, 198, does not change, but the atomic number of the new nucleus increases by one. The new atomic number is 80, which is mercury, Hg. Study Check 9.8 In an experimental treatment, a person is given boron-10, which is taken up by malignant tumors. When bombarded with neutrons, boron-10 decays by emitting alpha particles that destroy the surrounding tumor cells. Write the balanced nuclear equation for this experimental procedure.