MIME-Version: 1.0 Content-Type: multipart/related; boundary="----=_NextPart_01C5FCB6.62111020" This document is a Single File Web Page, also known as a Web Archive file. If you are seeing this message, your browser or editor doesn't support Web Archive files. Please download a browser that supports Web Archive, such as Microsoft Internet Explorer. ------=_NextPart_01C5FCB6.62111020 Content-Location: file:///C:/D9286902/Fall05.Phys3650.exam3.SolutionWeb.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" TIME OF COMPLETION_______________

TIME OF COMPLETION_______________  &nb= sp;         NAME____SOLUTION= _________________________

 

 

<= span style=3D'font-family:"Times New Roman"'>&nb= sp;            =             &nb= sp;            =      DEPARTMENT OF NATURAL SCIENCES

 

PHYS 3650, Exam 3            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;  Section 1

Version 1           &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =       December 5, 2005

Total Weight: 100 points

 

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

 

2.         = Authorized references include your calculator with calculator handbook, and the Refere= nce 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 all problems.  Show all work; part= ial 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:        &= nbsp;      10:30 a.m.

Stop:            &= nbsp;   11:45 a.m

 

 

 

&= nbsp;           &nbs= p; PROBLEM

 

&= nbsp;           &nbs= p;    POINTS

 

&= nbsp;           &nbs= p;   CREDIT

 

   &nbs= p;            &= nbsp;    1-6

 

   &nbs= p;            &= nbsp;    30

 

&= nbsp;

 

   &nbs= p;            &= nbsp;     7

 

   &nbs= p;            &= nbsp;    20

 

&= nbsp;

 

8

 

   &nbs= p;            &= nbsp;    20

 

&= nbsp;

 

   &nbs= p;            &= nbsp;     9

 

   &nbs= p;            &= nbsp;    15

 

&= nbsp;

 

&= nbsp;           &nbs= p;        10

 

&= nbsp;           &nbs= p;        15

 

 

 

&= nbsp;           &nbs= p;    TOTAL

 

&= nbsp;           &nbs= p;       100

 

 

 

 

 

&= nbsp;          PERCENTA= GE

 

 

 

 

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 nuclear reaction is said to be exothermic if the total kinetic energy is  

  &nbs= p;        

  &nbs= p;         a. Less after the reaction than before.

  &nbs= p;        

  &nbs= p;         b. Equal both before and after the reaction.

(5)  &= nbsp;   

  &nbs= p;         c. Greater after the reactio= n than before.

  &nbs= p;        

  &nbs= p;         d. Zero after the reaction.

 

 

 

 

2.   The number of protons in an atom is

 

  &nbs= p;         a.  Zero.

 

  &nbs= p;         b. Equal to the number of neutrons.

(5)

  &nbs= p;         c. Equal to the number of electrons.

 

  &nbs= p;         d. The same for all elements.

 

 

 

3. Which of the following statements concerning = the nuclear force is false?

 

  &nbs= p;         a. The nuclear force is very short-ranged.

 

  &nbs= p;         b. The nuclear force is very= weak and much smaller in relative magnitude than the electrostatic and gravitati= onal forces.

 

  &nbs= p;         c. The nuclear force is attractive and not repulsive.

(5)  &= nbsp;   

  &nbs= p;         d. The nuclear force acts on both protons and neutrons.

 

 

 

4.   = A gamma ray is also known as

 

  &nbs= p;         a. An electron. =

 

  &nbs= p;         b. A positron. <= /o:p>

(5)

  &nbs= p;         c. A helium nucleus. =

 

  &nbs= p;         d. A photon.

 

 

 

5.  During β+ decay

 

  &nbs= p;         a. A neutron is transformed to a proton.

 

  &nbs= p;         b. A proton is transformed t= o a neutron.

(5)

  &nbs= p;         c. A neutron is ejected from the nucleus.

 

  &nbs= p;         d. A proton is ejected from the nucleus.

 

 

 

 

6.  What happens to the half-life of a radioactive substance as it decays?

 

  &nbs= p;         a. It remains constant.

 

  &nbs= p;         b. It increases.

(5)

  &nbs= p;         c. It decreases.

 

  &nbs= p;         d. It could do any of these.

 

 

 

 

&nb= sp;

7.      = A   source is u= sed for 2.0 hours by a 75-kg student in a physics lab. Radioactive  decays by <= sub> decay with a half-life of 30 years. The average energy of the emitted betas is about 190= keV per decay. The  decay is qu= ickly followed by a  with an ene= rgy of 660 keV. Assuming the student absorbs all emitted energy, what effective dose (in rem) is received during lab?

 

 

l =3D ln(2) / T1/2 =3D ln(2) / (946080000 s) =3D= 7.33 x 10-10 s-1

        R =3D 1.00 x 10-6<= /sup> Ci  =3D 3.70 x 104 decays/s=

&nb= sp;

        Total # of decays over 2.00 hours =3D (3.70 x 10= 4 decays/s)(7200 s) =3D 2.66 x 108

 

        Energy delivered by alphas: =

 

        E =3D (= 2.66 x 108)(190 x 103 eV)=3D 5.06 x 1013 eV =3D 8.10 x 10-6 J

 

        AD =3D E/m =3D (8.10 x 10-6 J)/(75 kg) =3D 0.108 x 10-6 J/kg= =3D  0.108 x 10-4 rad

 

        AD(rem) =3D  AD(rad) X RBE

 

        AD =3D (0.108 x 10-4 rad= ) (1) =3D 0.108 x 10-4 rem

 

        Energy delivered by gammas:

 

        E =3D (= 2.66 x 108)(660 x 103 eV)=3D 1.76 x 1014 eV =3D 2.82 x 10-5 J

 

        AD =3D E/m =3D (2.82 x 10-5 J)/(75 kg) =3D 0.375 x 10-6 J/kg= =3D  0.375 x 10-4 rad

 

        AD(rem) =3D  AD(rad) X RBE

 

        AD =3D (0.375 x 10-4 rad= ) (1) =3D 0.375 x 10-4 rem

 

        Total:

 

         AD =3D 0.108 x 10-4 rem + 0.375 x 10-4 rem =3D= 0.483 x 10-4 rem =3D 4.83 mrem

 

8.  has a half-= life of 30.8 s.

a.  If we have  initially, = how many Cs nuclei are present?

        # of nuclei =3D # moles x NA =3D (8.80 x 10-6 g)/(124 g) x= NA =3D  = 4.27 x 1016

b. How many are present 2.0 min la= ter?

N =3D N0e- <= /span>l t

l =3D ln(2) / T1/2 =3D ln(2) / (30.8 s) =3D 0.02= 25 s-1

N =3D (4.27 x 10<= sup>16) e- (0.0225 s-1 ) (120 s) =3D  2.87 x = 1015

c. What is the activity at this ti= me?

R =3D l N  =3D (0.0225 s-1)(2.87 x 1015) =3D  6.= 45 x 1013 Bq =3D 1.74 x 103 Ci

d. After how much time will the activity drop to less than about 1 per second?

R =3D R0e- <= /span>l t

R/R0=3D e- <= /span>l t

ln(R/R0) =3D - <= span style=3D'font-size:12.0pt;line-height:200%;font-family:Symbol;color:black; mso-ansi-language:EN-US;mso-no-proof:no'>l t

t =3D ln(R/R0)/( - l) =3D l= n ((1 Bq)/(6.45 x 1013Bq= ))/(-0.0225 s-1) =3D 1413 s =3D 23.55 min

 

&n= bsp;

&n= bsp;

9.  The <= sub> isotope of hydrogen, which is called tritium (because it contains three nucleons), has a half-life of 12.33 yr. It can be used to measure the age of objects up to about 100 yr. It is produced in the upper atmosphere by cosmic rays and brought to Earth by rain. As an application, determine approximately the age of a bottle of wine whose  radiation is about  that presen= t in new wine.

l =3D ln(2) / T1/2 =3D ln(2) / (12.33 yr) =3D 0.= 0562 yr-1

R =3D R0e- <= /span>l t

R/R0=3D e- <= /span>l t

ln(R/R0) =3D - <= span style=3D'font-size:12.0pt;line-height:200%;font-family:Symbol;color:black; mso-ansi-language:EN-US;mso-no-proof:no'>l t

t =3D ln(R/R0)/( - l) =3D l= n (0.100)/(-0.0562 yr-1) =3D 41 yr

 

 

10.  and  The first f= ive decays of these two sequences are:

and

Determine the resulting intermedia= te daughter nuclei in case of and work out the sequence as a 5-point bonus.

&nb= sp;

232= 90Th  --> =   22888Ra=  + 42He  

&nb= sp;

228= 88Ra  --> =   22889Ac=  +  b- + (n)

 

228= 89Ac  --> =   22890Th=  +  b- + (n)

 

228= 90Th  --> =   22488Ra=  + 42He  

 

224= 88Ra  --> =   22086Rn=  + 42He  

 

***************************

235= 92U  --> <= span style=3D'mso-spacerun:yes'>  23190Th=  + 42He  

 

231= 90Th -->   <= /span>23191Pa  +  b- + (n)

 

231= 91Pa  --> =   22789Ac=  + 42He  

 

227= 89Ac -->   <= /span>22790Th  +  b- + (n)

 

227= 90Th  --> =   22389Ra=  + 42He  

 

 

 

 

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= 7

PHYS 3650 Exam 3, Version 1

Fall 2005

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