Lecture handout Ch21 with answers PDF

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Handout Chapter 21 Nuclear Chemistry LECTURE HANDOUT CHAPTER 21 The following problems were selected to help you apply the key topics outlined in the first lecture for this Chapter. They are presented along with the related paragraphs and learning goals they connect to from your text and lecture slides. While we will begin to work on these problems in class you should try to practice these concepts and answer these for practice. Chapter 21 Nuclear Chemistry

Handout Chapter 21 Nuclear Chemistry Question 1 Iodine-137 decays to give xenon-137, which decays to give cesium-137. What are the modes of decay in these two reactions? Answer 137 53 137 54

I®

Xe ®

137 54

137 55

Xe + –10b

Cs + –10b

Both processes are b decays. The effect of the b emission process is that the atomic number increases by 1, leaving the mass number unchanged.

Question 2. Write a balanced nuclear equation for: a. Beta emission by 28Mg c. Electron capture by 129Cs b. Alpha emission by 255Lr d. Positron emission by 25Al Answer The effect of the processes in this problem are that

 b emission increases the atomic number by 1, leaving the mass number unchanged;   emission reduces the atomic number by 2 and the mass number by 4;  electron capture reduces the atomic number by 1, leaving the mass number unchanged; and  positron emission reduces the atomic number by 1, leaving the mass number unchanged. (a) (b) (c)

28 12

Mg ®

b + 28 Al 13

0 –1

255 103

Lr ® 42 + 251 Md 101

129 55

Cs + –10b ®

129 54

Xe

(d) 13 Al ® 1b + 12Mg Electron capture and positron emission accomplish the same reduction of protons in the nucleus 25

0

25

Question 3. How can the belt of stability be used to predict the probable decay mode of an unstable nuclide? We are to describe how the belt of stability (shown in Figure 21.2) can be used to predict the possible decay modes of radioactive nuclides. The belt of stability in Figure 21.2 plots the number of neutrons versus the number of protons. If the nuclide lies in the belt of stability (green dots on the plot in Figure 21.2), it is not radioactive and is stable. If it lies above the belt of stability, it is neutron rich and tends to undergo b decay to increase the number of protons and reduce the number of neutrons in its nucleus. If it lies below the belt of stability, it is neutron poor and tends to undergo positron emission or electron capture to increase the number of neutrons and reduce the number of protons in its nucleus.

Handout Chapter 21 Nuclear Chemistry

From Figure 21.2 we can see that an element often has several radioactive isotopes (light gray dots).

Question 4. Explosions at a disabled nuclear power station in Fukushima, Japan, in 2011 may have released more cesium-137 (t1/2 = 30.2 years) into the ocean than any other single event. How long will it take the radioactivity of this radionuclide to decay to 5.0% of the level released in 2011? Answer For cesium-137 that was released at the Fukushima power plant in 2011 and that has a halflife of 30.2 yr, we are to calculate the time it will take for the radioactivity to decay to 5.0% of the level released. The time of decay of a radionuclide can be found from rearrangement of the equation Nt t/t =0.5 1 2 N0 ln

Nt t =–0.693 t1 2 N0 æ N ö t1 2 ´ ç ln t ÷ è N0 ø t= - 0.693

For N0 = 100 and Nt = 5.00, and t1/2 = 30.2 yr, the time needed for cesium-137 to decay to 5.0% is æ 5.0 ö 30.2 yr ´ç ln ÷ è 100 ø t= =131 yr - 0.693

This means that the year that the level of radioactivity will be 5% will be 2011 + 131 = 2142.

Handout Chapter 21 Nuclear Chemistry Question 5 Spent fuel removed from nuclear power stations contains plutonium-239 (t1/2 = 2.41 × 104 years). How long will it take a sample of this radionuclide to reach a level of radioactivity that is 2.5% of the level it had when it was removed from a reactor? Answer The time of decay of a radionuclide can be found from rearrangement of the equation Nt t/t =0.5 1 2 N0 ln

Nt t =–0.693 t1 2 N0 æ N ö t1 2 ´ ç ln t ÷ è N0 ø t= - 0.693

For N0 = 100 and Nt = 2.5, and t1/2 = 2.41 ´ 104 yr, the time needed for plutonium-239 to decay to 2.5% is 4

t=

æ

ö

2.5 ÷ ( 2.41´10 yr ) ´çè ln100 ø - 0.693

=1.3´105 yr

This means that the level of radioactivity will be 2.5% in 1300 centuries. Question 6 In the years immediately following the explosion at the Chernobyl nuclear power plant, the concentration of 90Sr in cow’s mil in southern Europe was slightly elevated. Some samples contained as much as 1.25 Bq/L of 90Sr radioactivity. The half-life of strodium-90 is 28.8 years. a. White a balanced nuclear equation describing the decay of 90Sr. b. How many atoms of 90Sr are in a 200 mL glass of mil with 1.25 Bq/L of 90Sr radioactivity? c. Why would strontium-90 be more concentrated in milk than other foods, such as grains, fruits and vegetables? Answer For the radioactive isotope 90Sr, we are to write a balanced nuclear equation corresponding to its β decay, calculate the atoms of 90Sr in 200 mL of milk that has 1.25 Bq/L of 90Sr radioactivity, and give a reason why 90Sr would be more concentrated in milk than in other foods.

Handout Chapter 21 Nuclear Chemistry Analyze (a) Beta decay increases the atomic number, leaving the mass number unchanged. (b) To calculate the atoms of 90Sr in the milk, we can use the equation milliliters of milk ´

1.25 Bq 1 disintegration/s =disintegrations per second ´ 1 Bq 1000 mL

That is the rate of the first-order decay of 90Sr that follows the rate law Rate =k [ 90 Sr]

where k = 0.693/t1/2. Given that the t1/2 for 90Sr is 28.8 yr, we need to convert from years to seconds. (a) 38 Sr ® –1 b + 39Y (b) The number of disintegrations in 200 mL of milk is 90

0

90

1.25 Bq 1 disintegration/s ´ = 0.250 disintegration/s 200 mL ´ 1000 mL 1 Bq

The rate constant k in reciprocal seconds is k=

0.693 1 yr 1d 1 hr 1 min ´ ´ ´ ´ = 7.63´10 –10 s –1 28.8 yr 365 d 24 hr 60 min 60 s

The concentration of 90Sr in the milk from the first-order rate law is é 90Srù = 0.250 disintegration/s =3.28 ´108 ë û 7.63 ´10–10 s –1

90

Sr atoms

(c) Strontium-90 is found in milk and not other foods because it is chemically similar to calcium, and milk is rich in calcium. Think About It In more familiar chemical concentration terms, the concentration of 90Sr in these samples is 3.28 ´108

1 mol 1 ´ =2.72 ´10–15 M Sr atoms ´ 6.022 ´1023 atoms 0.200 L

90...