13: Week Suggested Problems
Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is 34 8 9 19 E = hf = hc/ = (6.63 x 10 J-s)(3.00 x 10 m/s)(10 nm/m)/[(1.60 x 10 J/eV)]; 3 E = (1.24 x 10 eV-nm)/. A factor that appears in the analysis of electron energies is 9 2 2 19 2 28 ke2 = (9.00 x 10 N-m /C )(1.60 x 10 C) = 2.30 x 10 J-m = 1.44 eV-nm = 1.44 MeV-fm. ------------------------------------------------------------------------------------------------------------------P30.19 60 Co in an excited state emits a 1.33-MeV ray as it jumps to the ground state. What 27 is the mass (in u) of the excited cobalt atom? S30.19 The total rest energy of the excited atom includes the energy of the ray. m0c = (59.933820 u-c ) + [(1.33 MeV)/(931.5 MeV/u-c2)] = 59.93525 u-c . Therefore,
2 2 2
m0 = 59.93525 u
+ -
------------------------------------------------------------------------------------------------------------------P30.23 22 Na is radioactive. Is it a or emitter? Write down the decay reaction, and 11 estimate the maximum KE of the emitted .
22 S30.23 If 22 Na were a emitter, the resulting nucleus would be 12Mg , which has too few 11 + neutrons relative to the number of protons to be stable. Thus we have a emitter.
For the reaction 22 Na 22 Ne + + + , if we add 11 electrons to both sides in order to use atomic 11 10 masses, we see that we have two extra electron masses on the right. The kinetic energy of the will be maximum if an approximately zero-energy neutrino is emitted. If we ignore the recoil of the neon, the maximum kinetic energy is KE = [m( Na) m( Ne) 2m(e)]c = 1.82 MeV. ------------------------------------------------------------------------------------------------------------------P30.43 The iodine isotope 131 is used in hospitals for diagnosis of thyroid function. If 532 53 I g are ingested by a patient, determine the activity (a) immediately, (b) 1.0 h later when the thyroid is being tested, and (c) 6 months later. Use Appendix F.
22 22 2 2 2 +
= [(21.994434 u) (21.991383 u) 2(0.00054858 u)]c (931.5 MeV/u-c )
Due 18 April 2000
P108S00: Week 13 Homework
Page 1 of 4
S30.43 The decay constant is = 0.693/T = 0.693/(8.04 days)(24 h/day)(3600 s/h) = 9.976 x 10 s . The initial number of nuclei is N0 = [(532 x 10 g)/(131 g/mol)](6.02 x 10 atoms/mol) = 2.445 x 10 (a) When t = 0, we get N = N0 e
t 6 23 18 7 1
nuclei.
= (9.976 x 10 s )(2.445 x 10 ) e = 2.44 x 10
7 1 3
7 1
18
0
12
decays/s.
(b) When t = 1.0 h, the exponent is t = (9.976 x 10 s )(1.0 h)(3600 s/h) = 3.591 x 10 , so we get N = N0 e
t
= (9.976 x 10 s )(2.445 x 10 ) e
7 1
7 1
18
0.003591
= 2.43 x 10
12
decays/s.
(c) When t = 6 months, the exponent is t = (9.976 x 10 s )(6 mo)(30 days/mo)(24 h/day)(3600 s/h) = 15.51, so we get N = N0 e
t
= (9.976 x 10 s )(2.445 x 10 ) e
7 1
18
15.51
=
4.48 x 105 decays/s.
------------------------------------------------------------------------------------------------------------------P30.59 Determine the density of nuclear matter in kg/m , and show that it is essentially the same for all nuclei. (b) What would be the radius of the Earth if it had its actual mass but had the density of nuclei? (c) What would be the radius of a 238U nucleus if it had the density 92 of the Earth? S30.59 (a) The mass of a nucleus with mass number A is A u and its radius is r = (1.2 x 1015 m)A1/3. Thus the density is = m/V = A(1.66 x 10 (b)
27 3
kg/u)/[(4/3)r ] = A(1.66 x 10
3
27
kg/u)/[(4/3)(1.2 x 10
15
m) A]
3
= We find the radius from M = V;
2.29 x 1017 kg/m3,
independent of A.
(c)
24 17 3 5.98 x 10 kg = (2.29 x 10 kg/m3)[(4/3)R ], which gives R = 180 m. For equal densities, we have
= MEarth/[(4/3)REarth ] = mU/[(4/3)rU ]; (5.98 x 10 which gives
24
3
3
kg)/(6.38 x 10 m) = (238 u)(1.66 x 10
6
3
27
kg/u)/rU ,
3
rU =
2.58 x 1010 m.
-------------------------------------------------------------------------------------------------------------------
Due 18 April 2000
P108S00: Week 13 Homework
Page 2 of 4
P30.71 The nuclide 191Os decays with maximum energy of 0.14 MeV accompanied by 76 rays of energies 0.042 MeV and 0.129 MeV. (a) What is the daughter nucleus? (b) Draw and energy-level diagram showing the ground states of the parent and daughter and states excited of the daughter. To which of the daughter states does decay of 191Os occur? 76 S30.71 (a) We find the daughter nucleus by balancing the mass and charge numbers: Z(X) = Z(Os) Z(e ) = 76 ( 1) = 77; A(X) = A(Os) A(e ) = 191 0 = 191, 191 . so the daughter nucleus is 77Ir (b) In order to see the decays, the decay must sometimes go to the higher of the two excited states of
191 76 Os
(0.14 MeV) (0.042 MeV) (0.129 MeV)
191 77 Ir* 191 77 Ir* 191 77 Ir
the daughter. Because there is only one maximum energy, the decay must be to the higher excited state. -------------------------------------------------------------------------------------------------------------------
Week 13: Assigned Problems to Hand In
P30.26 When
23 10 Ne
(mass = 22.9945 u) decays
23 11
Na to (mass = 22.9898 u), what is the
maximum kinetic energy of the emitted electron? What is its minimum energy? What is the energy of the neutrino in either case? S30.26 The kinetic energy of the electron will be maximum if no neutrino is emitted. If we ignore the recoil of the sodium, the maximum kinetic energy of the electron is 23 23 KE = [m( Ne) m( Na)]c2 = [(22.994465 u) (22.989767 u)]c (931.5 MeV/uc ) =
2 2
4.38 MeV.
When the neutrino has all of the kinetic energy, the minimum KE of the electron is 0. The sum of the kinetic energy of the electron and the energy of the neutrino must be from the mass difference, so the energy range of the neutrino will be
0 E 4.38 MeV.
-----------------------------------------------------------------------------------------------------------------P30.36 Look up the half life of S30.36 The decay constant is = 0.693/T = 0.693/(4.468 x 10 yr)(3.16 x 10 s/yr) =
9 7
238 92
U in Appendix F, and then determine its decay constant.
4.91 x 1018 s1.
-----------------------------------------------------------------------------------------------------------------P30.56 An ancient club is found that contains 170 g of carbon and has an activity of 5.0 decays per second. Determing the age assuming that in living trees the ratio of 14C/ 12C atoms is about 1.3 x 10-12. Due 18 April 2000 P108S00: Week 13 Homework Page 3 of 4
S30.56 The decay constant is = 0.693/T = 0.693/(5730 yr) = 1.209 x 10 /yr. Because the fraction of atoms that are C is so small, we use the atomic weight of the number of carbon atoms in 170 g: N = [(170 g)/(12 g/mol)](6.02 x 10 atoms/mol) = 8.53 x 10 atoms, so the number of
14 23 24 14 12 4
C to find
C nuclei in a sample from a living tree is
12
N14 = (1.3 x 10
)(8.53 x 10 ) = 1.11 x 10
24
13
nuclei.
Because the carbon is being replenished in living trees, we assume that this number produced the activity when the club was made. (Actually, the average atmospheric ratio has changed with time in a way that has been measured.) We determine its age from t N = N14 e ; 5.0 decays/s = [(1.209 x 10 /yr)/(3.16 x 10 s/yr)](1.11 x 10 This gives t = 1.8 x 10 yr.
4 4 7 13
nuclei) e (1.209
10
4
/ yr) t
------------------------------------------------------------------------------------------------------------------P30.66 When water is placed near an intense neutron source, the neutrons can be slowed down by collisions with water molecules and eventually captured by a hydrogen nucleus to 2 form the stable isotope called deuterium, 1H, giving off a gamma ray. What is the energy of the gamma ray? S30.66 The capture is 1H + 1 n 2H + . 1 0 1
1 1 2
The overall energy conservation equation is
m0( H) + m0( n) + KEH + HE n = m0( H) + 0 + KE + KED Because the kinetic energies of the particles are small, the gamma energy is approximately the energy released: Q = [m( 1H) + m( 1 n) m( 2H)]c2 = [(1.007825 u) + (1.008665 u) (2.014102 u)]c2(931.5 MeV/uc 2) = 2.22 MeV.
KE = Q =
2.22 eV
------------------------------------------------------------------------------------------------------------------P31.6 Calculate the energy released (or energy input required) for the reaction 9 12 4 Be (,n) 6 C. S31.6 For the reaction 9Be( , n)12C , we determine the Q-value: 4 6 Q = [m( 9Be) + m( 4He) m(n) m( 12C)]c2 = [(9.012182 u) + (4.002602 u) (1.008665 u) (12.000000 u)]c2(931.5 MeV/uc 2) = + 5.700 MeV.
Thus
5.700 MeV is released.
----------------------------------------------------------------------------------------------------------------Due 18 April 2000 P108S00: Week 13 Homework Page 4 of 4