(Answer) CHMY 161 Chapter 5 Homework

Question 1

The cesium isotope 13155Cs is used in cancer treatment. 13155Cs contains _____ protons.

55

 

76

 

131

 

 

186

 

Question 2

The cesium isotope 13155Cs is used in cancer treatment. 13155Cs contains _____ neutrons.

 

55

 

 

76

 

 

131

 

 

186

 

Question 3

Which type of radiation penetrates farthest into skin and body tissues:

 

alpha particles

 

 

beta particles

 

 

gamma radiation

Question 4

Identify the following nuclear equation as alpha decay, beta decay, positron emission, or gamma emission:

12755Cs → 12754Xe + 0+1e

 

alpha decay

 

 

beta decay

 

 

positron emission

 

 

gamma emission

 

Question 5

Identify the following nuclear equation as alpha decay, beta decay, positron emission, or gamma emission:

21885At → 21483Bi + 42He

 

alpha decay

 

 

beta decay

 

 

positron emission

 

 

gamma emission

 

 

Question 6

In the following nuclear equation, the unidentified type of radiation represented by a question mark is:

2813Al → 2814Si + ?

 

42He

 

0-1e

 

0+1e

 

gamma ray

 

10n

 

Question 7

In the following nuclear equation, the unidentified portion represented by a question mark is:

? + 5927Co → 5625Mn + 42He

 

42He

 

0-1e

 

0+1e

 

11H

 

10n

 

Question 8

In the following nuclear equation, the unidentified portion represented by a question mark is:

? → 147N + 0-1e

 

146N

 

146C

 

147N

 

147C

 

0+1e

 

Question 9

In the following nuclear equation, the unidentified portion represented by a question mark is:

0-1e + 7636Kr → ?

 

7635Br

 

7635Kr

 

7636Br

 

7636Kr

 

0+1e

 

Question 10

In the following nuclear equation, the unidentified portion represented by a question mark is:

42He + 24195Am → ? + 2(10n)

 

24395Am

 

24395Bk

 

24397Am

 

24497Bk

 

24397Bk

 

Question 11

The balanced nuclear equation for the alpha decay of the radioactive isotope platinum-190 is:

 

19078Pt → 18676Os + 42He

 

78190Pt → 76186Os + 24He

 

19078Pt → 18679Au + 0-1e

 

19081Pt → 18679Au + 42He

 

19078Pt + 42He → 19481Tl + 0-1e

 

Question 12

In the following nuclear equation, the unidentified type of radiation represented by a question mark is:

116C → 115B + ?

 

42He

 

0-1e

 

0+1e

 

gamma ray

 

 

11H

 

10n

 

Question 13

In the following bombardment reaction, the unidentified portion represented by a question mark is:

? + 4018Ar → 4319K + 11H

 

42He

 

0-1e

 

0+1e

 

gamma ray

 

 

11H

 

10n

 

Question 14

 

In the following nuclear equation, the unidentified portion represented by a question mark is:

188O + 24998Cf → ? + 4(10n)

 

26390Th

 

26690Th

 

227106Sg

 

263106Sg

 

266106Sg

 

Question 15

The balanced nuclear equation for the decay of polonium-210 to give lead-206 is:

21084Pb → 20682Po + 42He

 

20682Pb → 21084Po + 42He

 

21084Po + 42He → 20682Pb

 

20682Pb + 42He → 21084Po

 

21084Po → 20682Pb + 42He

 

20682Po + 42He → 21084Pb

 

Question 16

A radioisotope emits a positron to form titanium-48. The balanced nuclear equation is:

 

4823V → 4822Ti + 0+1e

 

4822Ti + 0+1e → 4823V

 

4823V + 0+1e → 4822Ti

 

4822Ti → 4823V + 0+1e

 

4823Ti → 4822V + 0+1e

 

4822V + 0+1e → 4823Ti

 

Question 17

The unit of measurement mrem is used to measure the property:

 

activity

 

 

absorbed dose

 

 

biological damage

 

Question 18

The unit of measurement microCi is used to measure the property:

 

activity

 

absorbed dose

 

biological damage

 

Question 19

The unit of measurement mrad is used to measure the property:

 

activity

 

absorbed dose

 

biological damage

Question 20

The activity of C-14 in a 70. kg human body is estimated to be 3.7 kBq. What is this activity in microcuries?

 

0.010 microCi

 

0.10 microCi

 

1.0 microCi

 

3.7 x 109 microCi

 

3.7 x 1010 microCi

 

3.7 x 1011 microCi

 

Question 21

Two samples of a radioisotope were spilled in a nuclear laboratory. The activity of one sample was 8 kBq and the other 15 microCi. Which sample produced the higher amount of radiation?

 

the sample with an activity of 8 kBq

 

 

the sample with an activity of 15 microCi

 

Question 22

A sample of F-18 has a half-life of 110 minutes. How many half-lives have elapsed after 330 minutes?

 

one half-life

 

 

two half-lives

 

 

three half-lives

 

Question 23

Cesium-137, a beta emitter, has a half-life of 30 y. How many milligrams of a 16 mg sample of cesium-137 would remain after 90 y?

 

0.13 mg

 

 

0.25 mg

 

 

0.5 mg

 

 

1.0 mg

 

 

2.0 mg

 

 

4.0 mg

Question 24

P-32, a radioisotope used to treat leukemia, has a half-life of 14.3 days. If a sample contains 8.0 mg of P-32, how many milligrams of P-32 remain after 42.9 days?

 

0.50 mg

 

 

1.0 mg

 

 

2.0 mg

 

 

3.0 mg

 

 

4.0 mg

 

Question 25

A sample of sodium-24 with an activity of 12 mCi is used to study the rate of blood flow in the circulatory system. If sodium-24 has a half-life of 15 h, calculate the activity of the sodium in mCi after 2.5 days.

 

0.38 mCi

 

 

0.75 mCi

 

 

1.5 mCi

 

 

3.0 mCi

 

 

6.0 mCi

 

 

12 mCi

Question 26

A 16 microgram sample of sodium-24 decays to 2.0 micrograms in 45 h. What is the half-life, in hours, of sodium-24?

 

5.6 h / half-life

 

 

11 h / half-life

 

 

15 h / half-life

 

 

23 h / half-life

Question 27

Cesium-137, a beta emitter, has a half-life of 30 y. How many years are required for 28 mg of cesium-137 to decay to 3.5 mg of cesium-137?

 

90 years

 

 

120 y

 

 

150 y

 

 

180 y

 

 

210 y

 

 

240 y

 

Question 28

A technician was accidentally exposed to potassium-42 while doing some brain scans for possible tumors. The error was not discovered until 36 h later when the activity of the potassium-42 sample was 2.0 microCi. If potassium-42 has a half-life of 12 h, what was the activity of the sample at the time the technician was exposed?

 

0.25 microCi

 

 

0.50 microCi

 

 

1.0 microCi

 

 

16 microCi

 

Question 29

The dosage of technetium-99m for a lung scan is 20. microCi/kg of body mass. How many millicuries of technetium-99m are needed for a 50.0 kg patient?

 

0.0025 mCi

 

 

0.4 mCi

 

 

1.0 mCi

 

 

2.5 mCi

 

 

1000 mCi

 

 

2500 mCi

 

Question 30

In a diagnostic test of leukemia, a patient receives 4.0 mL of a solution containing selenium-75. If the activity of the selenium-75 is 45 microCi/mL, what is the dose received by the patient?

 

0.089 microCi

 

11 microCi

 

180 microCi

 

13 500 microCi

 

4.0 mL solution * (45 microCi / 1 mL solution) = 180 microCi of selenium-75

 

Question 31

1 / 1 pts

A vial contains radioactive iodine-131 with an activity of 2.0 mCi/mL. If a thyroid test requires 3.0 mCi in an “atomic cocktail,” how many milliliters are used to prepare the iodine-131 solution?

 

1.5 x 10-3 mL

 

1.5 mL

 

1.5 x 103 mL

 

6.0 x 10-3 mL

 

6.0 mL

 

Question 32

Fluoride-18, which has a half-life of 110 min, is used in PET scans. If 100. mg of fluoride-18 are shipped at 8:00 am, how many milligrams of the radioisotope are still active if the sample arrives at the radiology laboratory at 1:30 pm?

 

3.13 mg

 

 

6.25 mg

 

 

12.5 mg

 

 

25.0 mg

 

 

50.0 mg

 

 

  1. mg

Question 33

If the amount of radioactive iodine-123 in a sample decreases from 0.4 g to 0.1 g in 26.4 h, what is the half-life, in hours, of iodine-123?

 

6.60 h/half-life

 

 

13.2 h/half-life

 

 

26.4 h/half-life

 

 

39.6 h/half-life

 

 

52.8 h/half-life

 

Question 34

The half-life of oxygen-15 is 124 s. If a sample of oxygen-15 has an activity of 4000. Bq, how many minutes will elapse before it reaches an activity of 500. Bq?

 

2.07 min

 

 

6.20 min

 

 

8.27 min

 

 

372 min

 

22 320 min

Question 35

In a fission reaction, uranium-235 is bombarded with a neutron producing strontium-94, another nucleus, and 3 neutrons. The balanced equation for the fission reaction is:

10n + 23592U → 9438Xe + 13954Sr + 3(10n) + energy

 

10n + 23592U → 9454Sr + 13938Xe + 3(10n) + energy

 

9438Sr + 13954Xe + 3(10n) + energy → 10n + 23592U

 

10n + 23592U → 9438Sr + 13954Xe + 3(10n) + energy

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