PH 103
5.
Practice Problems - Chapter 30
May, 2004
(a) What is the fraction of the hydrogen atom's mass that is in the nucleus? (b) What is the fraction
of the atom's volume that is occupied by the nucleus? (c) What is the density of nuclear matter?
Compare this with water.
(a) The fraction of mass is
mp/(mp + me) = (1.6710–27 kg)/(1.6710–27 kg + 9.1110–31 kg) =
0.99945.
(b) The fraction of volume is
(rnucleus/ratom)3 = [(1.210–15 m)/(0.5310–10 m)]3 =
1.210–14.
(c) If we take the density of the hydrogen nucleus as the density of nuclear matter, we get
 = mp/(4/3)r3 = (1.6710–27 kg)/ (4/3) (1.210–15 m)3 = 2.31017 kg/m3.
The density of water is 1000 kg/m3, so nuclear matter is 1014  greater.
28.
The isotope
218
84P
can decay by either α or β- emission. What is the energy release in each case?
  82Pb + 2He . The Q value is
For alpha decay we have 84Po
218
214
Q = [m( Po) – m( Pb) – m(4He)]c2
= [(218.008965 u) – (213.999798 u) – (4.002602 u)]c2(931.5 MeV/uc2) =
218
0
For beta decay we have 218
  85At + –1e . The Q value is
84Po
Q = [m(218Po) – m(218At)]c2
= [(218.008965 u) – (218.00868 u)]c2(931.5 MeV/uc2) =
0.27 MeV.
218
43.
214
4
6.12 MeV.
The iodine isotope 13153I is used in hospitals for diagnosis of thyroid function.
(a) If 532 µg are ingested by a patient, determine the activity immediately. (b) What is the activity
1.0 h later when the thyroid is being tested? (c) What is the activity 6 months later?
The decay constant is
 = 0.693/T1/2 = 0.693/(8.04 days)(24 h/day)(3600 s/h) = 9.976 10–7 s–1.
The initial number of nuclei is
N0 = [(53210–6 g)/(131 g/mol)](6.021023 atoms/mol) = 2.4451018 nuclei.
(a) When t = 0, we get
N = N0 e – t = (9.97610–7 s–1)(2.4451018) e 0 = 2.441012 decays/s.
(b) When t = 1.0 h, the exponent is
t = (9.97610–7 s–1)(1.0 h)(3600 s/h) = 3.59110–3,
so we get
N = N0 e – t = (9.97610–7 s–1)(2.4451018) e – 0.003591 = 2.431012 decays/s.
(c) When t = 6 months, the exponent is
t = (9.97610–7 s–1)(6 mo)(30 days/mo)(24 h/day)(3600 s/h) = 15.51,
so we get
N = N0 e – t = (9.97610–7 s–1)(2.4451018) e – 15.51 = 4.48105 decays/s.
66.
When water is placed near an intense neutron source, the neutrons can be slowed down by
collisions with the water molecules and eventually captured by a hydrogen nucleus to form the
stable isotope called deuterium, 21H, giving off a gamma ray. What is the energy of the gamma
ray?
The capture is 1H + 10 n  1H +  .
Because the kinetic energies of the particles are small, the gamma energy is the energy released:
Q = [m(1H) + m(1n) – m(2H)]c2
= [(1.007825 u) + (1.008665 u) – (2.014102 u)]c2(931.5 MeV/uc2) =
2.22 MeV.
1
2
PH 103
10.
Practice Problems - Chapter 31
Calculate the Q-value for the “capture” reaction
16
8O
(α, γ) 2010Ne.
For the reaction 168O (,  )20
10Ne , we determine the Q-value:
Q = [m(16O) + m(4He) – m(20Ne)]c2
= [(15.994915 u) + (4.002602 u) – (19.992435 u)]c2(931.5 MeV/uc2) =
24.
May, 2004
+ 4.734 MeV.
What is the average kinetic energy of protons at the center of a star where the temperature is 107 K?
We find the average kinetic energy from
KE = (3/2) kT = (3/2) (1.3810–23 J/K)(107 K )/(1.6010–19 J/eV) = 1.29103 eV =
40.
1.3 keV.
A 0.018-Ci sample of 3215P is injected into an animal for tracer studies. If a Geiger counter
intercepts 20% of the emitted β particles, what will be the counting rate, assumed 90%?
If the counter counts 90% of the intercepted  particles, we have
n = (0.90)(0.20)(0.01810–6 Ci)(3.71010 decays/s · Ci) = 120 counts/s.
45.
Huge amounts of radioactive 13153I were released in the accident at Chernobyl in 1986. Chemically,
iodine goes to the human thyroid. (Doctors can use it for diagnosis and treatment of thyroid
problems.) In a normal thyroid, 13153I absorption can cause damage to the thyroid. (a) Write down
the decay scheme for 13153I (see page 1066 for the decay mode). (b) Its half-life is 8.04 d; how long
would it take for ingested 13153I to become 10% of the initial value? (c) Absorbing 1 mCi of 13153I
can be harmful; what mass of iodine is this?
(a) 53I  54Xe + – 01e +  +  .
(b) We find the number of half-lives from
N/N0 = (½)n;
(0.10) = (½)n, or n log 2 = log 10, which gives n = 3.32.
Thus the elapsed time is
∆t = nT1/2 = (3.32)(8.0 days) = 27 days.
(c) We find the number of atoms from
Activity = N;
(1.0010–3 Ci)(3.71010 decays/s · Ci) = [0.693/(8.0 days)(86,400 s/day)]N,
which gives N = 3.691013 atoms.
The mass is
m = [(3.691013 atoms)/(6.021023 atoms/mol)](131 g/mol) =
8.010–9 g.
131
131
PH 103
4.
Practice Problems - Chapter 32
May, 2004
What is the time for one complete revolution for a very high-energy proton in the 1.0 km radius
Fermilab accelerator?
Very high-energy protons will have a speed v  c. Thus the time for one revolution is
t = 2πr/v = 2π(1.0103 m)/(3.00108 m/s) = 2.110–5 s =
21 µs.
19.
How much energy is released in the decay
+
µ+ +
µ?
For the reaction π+  µ+ + µ , we determine the Q-value:
Q = [m(π+) – m(µ+)]c2
= [(139.6 MeV/c2) – (105.7 MeV/c2)]c2 = 33.9 MeV.
Thus
33.9 MeV
is released.
25.
What are the wavelengths of the two photons produced when a proton and antiproton at rest
annihilate?
The energy of the two photons must be the rest mass energy of the proton and antiproton:
2m0c2 = 2hf = 2hc/;
2(938.3 MeV/c2)c2 = 2(1.2410–12 MeV · m)/, which gives  =
39.
1.3210–15 m.
What particles do the following quark combinations produce? Note: underline means the
antiparticle. Refer to Table 32-2 and Table 32-4. (a) u u d, (b) u u s, (c) u s, (d) d u, (e) c s.
(a) For u u d we have charge +1, baryon number = + 1, while strangeness, charm, bottomness, and
topness = 0. Thus we have
p.
(b) For u us = u u s we have charge = – 1, baryon number = – 1, strangeness = + 1, while charm,
bottomness, and topness = 0. Thus we have + .
(c) For u s we have charge = – 1, baryon number = 0, strangeness = – 1, while charm, bottomness, and
topness = 0. Thus we have K–.
(d) For d u we have charge = – 1, baryon number = 0, strangeness = 0, while charm, bottomness, and
topness = 0. Thus we have π–.
(e) For cs we have charge = – 1, baryon number = 0, strangeness = – 1, charm = – 1, while bottomness,
and topness = 0. Thus we have D –S .