Problem#1
(a) Calculate the minimum energy required to remove one neutron from the nucleus 17O8. This is called the neutronremoval energy. (See Problem 43.52.) (b) How does the neutronremoval energy for 17O8 compare to the binding energy per nucleon for 17O8, calculated using Eq. (43.10)?
Answer
The minimum energy to remove a proton or a neutron from the nucleus is equal to the energy
difference between the two states of the nucleus, before and after removal.
17O8 → 1n0 + 16O8
∆m = m(1n0) + m(16O8)
The electron masses cancel when neutral atom masses are used.
∆m = 1.008665 u + 15.994915 u – 16.999132 u = 0.004448 u.
The energy equivalent of this mass increase is (0.004448 u)(931.5 MeV/u) = 4.14 MeV.
(b) Following the same procedure as in part (a) gives
∆M = 8MH + 9Mn – 17M8 = 8(1.007825 u) + 9(1.008665 u) – 16.999132u = 0.1415u.
EB = (0.1415 u)(931.5 MeV/u) = 131.8 MeV.
EA/A = 7.75 MeV/nucleon
Problem#2
The neutral atomic mass of 14C6 is 14.003242 u. Calculate the proton removal energy and the neutron removal energy for 15N7. (See Problems 43.52 and 43.53.) What is the percentage difference between these two energies, and which is larger?
Answer;
The minimum energy to remove a proton or a neutron from the nucleus is equal to the energy
difference between the two states of the nucleus, before and after removal.
proton removal:
7N15 = 1H1 + 6C14
∆m = m(1H1) + m(6C14) – m(7N15)
The electron masses cancel when neutral atom masses are used.
∆m = 1.007825 u +14.003242 u – 15.000109 u = 0.01096 u.
The proton removal energy is
E = 0.01096u x 931.5 MeV/u = 10.2 MeV
Neutron remova: 7N15 = 0n1 + 7N14
∆m = m(0n1) + m(7N14) – m(7N15)
The electron masses cancel when neutral atom masses are used.
∆m = 1.008665 u +14.003074 u – 15.000109 u = 0.01163 u.
The removal energi is 0.01163u x 931.5 MeV/u = 10.8 MeV
Problem#3
Radioactive Fallout. One of the problems of in-air testing of nuclear weapons (or, even worse, the use of such weapons!) is the danger of radioactive fallout. One of the most problematic nuclides in such fallout is strontium-90 (90Sr), which breaks down by β- decay with a half-life of 28 years. It is chemically similar to calcium and therefore can be incorporated into bones and teeth, where, due to its rather long half-life, it remains for years as an internal source of radiation. (a) What is the daughter nucleus of the 90Sr decay? (b) What percentage of the original level of 90Sr is left after 56 years? (c) How long would you have to wait for the original level to be reduced to 6.25% of its original value?
Answer:
Use the decay scheme and half-life of 90Sr to find out the product of its decay and the amount left after a given time.
The particle emitted in β-1 decay is an electron, -1e0. In a time of one half-life, the number of radioactive nuclei decreases by a factor of 2.
6.25% = 1/16 = 2-4
(a) 38Sr90 → -1e0 + 39Y90. The daughter nucleus is 39Y90.
(b) 56 y is 2T1/2so N = N0/22 = N0/4; 25% is left
(c) N/N0 = 2-n
N/N0 = 6.25% = 1/16 = 2-4
So, t = 4T1/2= 112 y
Problem#4
Thorium 230Th90 decays to radium 226Ra88 by α emission. The masses of the neutral atoms are 230.033127 u for 230Th90 and 226.025403 u for 226Ra88. If the parent thorium nucleus is at rest, what is the kinetic energy of the emitted α particle? (Be sure to account for the recoil of the daughter nucleus.)
Answer:
1 u is equivalent to 931.5 MeV
The α-particle will have 226/230 of the released energy
226(mTh – mRa – mα)/230 = 5.032 x 10-5u or 4.69 MeV
Problem#5
The atomic mass of 25Mg12 is 24.985837 u, and the atomic mass 25Al13 of is 24.990429 u. (a) Which of these nuclei will decay into the other? (b) What type of decay will occur? Explain how you determined this. (c) How much energy (in MeV) is released in the decay?
Answer:
(a) The heavier nucleus will decay into the lighter one.
13Al25 will decay into 12Mg25.
(b) Determine the emitted particle by balancing A and Z in the decay reaction
This gives 13Al25 → 12Mg25 + +1e0.
The emitted particle must have charge +e and its nucleon number must be zero. Therefore, it is a β+ particle, a positron.
(c) Calculate the energy defect ΔM for the reaction and find the energy equivalent of . ΔM Use the nuclear masses for 13Al25 and 12Mg25, to avoid confusion in including the correct number of electrons if neutral atom masses are used
The nuclear mass for 13Al25 is Mnuc(13Al25) = 24.990429 u – 13(0.000548580 u) = 24.983297 u
The nuclear mass for 12Mg25 is Mnuc(12Mg25) = 24.985837 u – 12(0.000548580 u) = 24.979254 u
The mass defect for the reaction is
∆M = Mnuc(13Al25) – Mnuc(12Mg25) – M(+1e0)
∆M = 24.983297 u – 24.979254 u – 0.00054858 u = 0.003494 u
Q = ∆Mc2 = 0.003494 u x 931.5 MeV/u = 3.255 MeV
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