TIME OF COMPLETION_______________           NAME___SOLUTION__________________________

 

 

                                           DEPARTMENT OF NATURAL SCIENCES

 

PHYS 3650, Exam 3                                                                                                             Section 1

Version 1                                                                                                                 December 1, 2003

Total Weight: 100 points

 

1.         Check your examination for completeness prior to starting.  There are a total of ten (10) problems on seven (7) pages.

 

2.         Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet  (provided by your instructor).

 

3.         You will have 75 minutes to complete the examination.

 

4.         The total weight of the examination is 100 points.

 

5.         There are six (6) multiple choice and four (4) calculation problems. Work five (5) multiple choice problems and four (4) calculation problems.  Show all work; partial credit will be given for correct work shown.

 

6.         If you have any questions during the examination, see your instructor who will

be located in the classroom.

 

7.         Start:                10:30 a.m.

Stop:                11:45 a.m

 

 

 

               PROBLEM

 

                 POINTS

 

                 CREDIT

 

                     1-6

 

                      30

 

 

 

                       7

 

                      15

 

 

 

8

 

                      20

 

 

 

                       9

 

                      15

 

 

 

                      10

 

                      20

 

 

 

                 TOTAL

 

                     100

 

 

 

 

 

           PERCENTAGE

 

 

 

CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT.

 

 

 

 

 

  1. A nucleus is specified completely by

 

    1. Its atomic number.

 

    1. Its proton number.

(6)

    1. The number of nucleons it has.

 

    1. Giving both A and Z.

 

 

 

  1. When a nucleus undergoes alpha decay, the daughter as compared to the parent has

 

    1. The same Z and an N reduced by 4.

 

    1. Z reduced by 4 and A reduced by 2.

(6)

    1. A reduced by 4 and Z reduced by 2.

 

    1. Z increased by 2 and N reduced by 2.

 

 

 

  1. When a nucleus undergoes electron beta decay, the daughter as compared to the parent has

 

    1. The same Z and an N reduced by 1.

(6)

    1. Z reduced by 1 and A reduced by 1.

 

    1. A reduced by 1 and Z increased by 1.

 

    1. None of these.

 

 

 

  1. The mass of the nucleus composed of several nucleons is

 

    1. Always less than the sum of masses of its constituents.

 

    1. Sometimes less than the sum of masses of its constituents.

(6)

    1. Always more than the sum of masses of its constituents.

 

    1. Always equal to the sum of masses of its constituents.

 

 

 

  1.  An antineutrino always accompanies

 

    1. a decay.

 

    1. b-- decay.

(6)

    1. Gamma decay.

 

    1. Neutron emission.

 

 

 

  1. Given that a freshly prepared radioactive isotope has a half-life of 10 days, the percentage of it remaining after 30 days is

 

    1. 30.0 %.

 

    1. 10.0 %.

(6)

    1. 12.5 %.

 

    1. 72.5 %.

 

 

  1. A sample of radioactive isotope is found to have an activity of 115.0 Bq immediately after it is pulled from the reactor that formed it. Its activity 2 hours and 15 minutes later is measured to be 85.2 Bq.

 

    1. What is the decay constant of the isotope?

 

R = Roe-l t

 

R / Ro  = e-l t

 

ln (R / Ro)  = -l t

 

l = - ln (R / Ro)  / t

 

l = - ln (85.2 Bq/ 115.0 Bq)  / (8100 s) = 3.703 x 10-5 s-1

 

    1. What is the half-life of the sample?

 

T1/2 = ln (2) / l  = 18,718 s

 

    1. How many radioactive nuclei were there in the sample initially?

  

 

R0 = l N0

 

N0 = R0 /l = (115.0 Bq) / (3.703 x 10-5 s-1) = 3.11 x 106

 

  1. The rubidium isotope 87Rb is a b emitter with a half-life of 4.90 x 1010 years that decays into 87Sr. It is used to determine the age of rocks and fossils. Rocks containing the fossils of early animals contain a ratio of  87Sr to 87Rb of 0.0100. Assuming that there was no 87Sr present when the rocks were formed, calculate the age of these fossils.

           

l  = ln (2) / T1/2  

 

l  = 4.49 x 10-19 s-1

 

(N0 – N) / N = 0.0100

 

N0 / N = 1.0100

 

N / N0 = 0.9900990099

 

N = Noe-l t

 

N / No  = e-l t

 

ln (N / No)  = -l t

 

t = ln (N / No)  / (-l)

 

t = 2.22 x 1016 s = 7.04 x 108 years

 

 

 

  1. Alpha particles with an RBE of 13 deliver a 32-mrad whole-body radiation to a 66-kg patient.

 

    1. What dosage, in rem, does the patient receive?

 

 

Dose (rem) = dose (rad) x RBE

 

Dose (rem) = (32.0 x 10-3 rad) x 13 = 0.416 rem

 

 

    1. How much energy is absorbed by the patient?

  

Dose (rad) = Energy  / m

 

Energy = dose (rad) x m 

 

Energy = (32.0 x 10-3  10-2 J/kg) (66.0 kg) = 0.0211 J  

 

 

  1.  As part of a treatment program, a patient ingests a radioactive pharmaceutical containing 3215P, which emits b rays with an RBE of 1.50. The half-life of 3215P is 14.28 days, and the initial activity of the medication is 1.34 MBq.

 

    1. How many electrons are emitted over the period of 1 week?

 

l  = ln (2) / T1/2  

 

l  = 5.61 x 10-7 s-1

 

R0 = l N0

 

N0 = R0 /l = (1.34 x 106 Bq) / (5.61 x 10-7 s-1) = 2.39 x 1012  nuclei

 

N = Noe-l t

 

N = (2.39 x 10 12) e-(5.61 x 10-7 s-1) (7 x 24 x 3600 s) = 1.70 x 1012 nuclei

 

 

#of electrons emitted = N0 –N = 6.88 x 1011

 

    1. If the b rays have an energy of 705 keV, what is the total amount of energy absorbed by the patient’s body in one week?

 

TE = (6.88 x 1011) (705000 eV) = 4.850 x 1017 eV = 7.760 x 10-2 J

 

 

    1. Find the absorbed dosage in rem, assuming the radiation is absorbed by 125 g of tissue. 

 

Dose (rad) = (7.760 x 10-2 ) J/ (0.125 kg) = 62.08 rad

 

Dose (rem) = (62.08 rad) x (1.50) = 93.12 rem