Problem#1
Calculate (a) the total binding energy and (b) the binding energy per nucleon of 12C. (c) What percent of the rest mass of this nucleus is its total binding energy?
Answer:
The binding energy of the nucleus is the energy of its constituent particles minus the energy of
the carbon-12 nucleus.
(a) In terms of the masses of the particles involved, the binding energy is
EB = (6mH + 6mn – mC-12)c2
Then, we get
EB = [6(1.007825 u) + 6(1.008665 u) – 12.000000u)](931.5 MeV/u) = 92.16 MeV
(b) The binding energy per nucleon is (92.16 MeV)/(12 nucleons) = 7.680 MeV/nucleon
(c) The energy of the C-12 nucleus is (12.0000 u)(931.5 MeV/u) = 11178 MeV. Therefore the percent of the mass that is binding energy is
92.16 MeV/11178 MeV = 0.8245%
Problem#2
A photon with a wavelength of 3.50 x 10-13 m strikes a deuteron, splitting it into a proton and a neutron. (a) Calculate the kinetic energy released in this interaction. (b) Assuming the two particles share the energy equally, and taking their masses to be 1.00 u, calculate their speeds after the photodisintegration.
Answer:
Conservation of energy tells us that the initial energy (photon plus deuteron) is equal to the
energy after the split (kinetic energy plus energy of the proton and neutron). Therefore the kinetic energy released is equal to the energy of the photon minus the binding energy of the deuteron.
The binding energy of a deuteron is 2.224 MeV and the energy of the photon is E = hc/λ. Kinetic energy is K = ½ mv2.
(a) The energy of the photon is
Eph = hc/λ = (6.62 x 10-34 Js)(2.998 x 108 m/s)/(3.50 x 10-13 m)
Eph = 5.68 x 10-13 J = 3.546 MeV
The binding of the deuteron is EB = 2.224 MeV
Therefore the kinetic energy is
K = Eph – EB = 3.546 MeV – 2.224 MeV = 1.321 MeV
(b) The particles share the energy equally, so each gets half. Solving the kinetic energy for v gives
K = ½ mv2
½ x 1.321 x 106 eV x 1.602 x 10-19 J/eV = ½ (1.6605 x 10-27 kg)v2
v2 = (2.116 x 10-13)/(1.6605 x 10-27) m2/s2
v = 1.13 x 107 m/s
Problem#3
Calculate the mass defect, the binding energy (in MeV), and the binding energy per nucleon of (a) the nitrogen nucleus, 14N7, and (b) the helium nucleus, 4He2 (c) How does the binding energy per nucleon compare for these two nuclei?
Answer:
The mass defect is the total mass of the constituents minus the mass of the atom.
1 u is equivalent to 931.5 MeV. 14N7 has 7 protons and 7 neutrons. 4He2 has 2 protons and
2 neutrons.
(a) the nitrogen nucleus, 14N7, is
∆E = ∆mc2 = (7mn + 7mH – mN)c2
∆E = [7(1.008665 u) + 7(1.007825 u) - 14.003074u]c2 = 105 MeV
and
the binding energy per nucleon is
∆E/nucleon = 105 MeV/14 nucleon = 7.48 MeV per nucleon
(b) the helium nucleus, 4He2
∆E = ∆mc2 = (2mn + 2mH – mHe)c2
∆E = [7(1.008665 u) + 7(1.007825 u) - 4.002603u]c2 = 28.3 MeV
and
the binding energy per nucleon is
∆E/nucleon = 28.3 MeV/4 nucleon = 7.07 MeV per nucleon
Problem#4
Calculate the binding energy per nucleon for the nuclei 86Kr36 and 180Ta73. Do your results confirm what is shown in Fig. 1—that for A greater than 62 the binding energy per nucleon deceases as A increases?
Answer:
86Kr36: A = 86 and Z = 36. N = A – Z = 50, which is even, so for the last term in Equation the total estimated binding energy is the sum of these five terms:
EB = C1A – C2A2/3 – C3Z(Z – 1)/A1/3 – C4(A – 2Z)2/A + C5Z-4/3
we use the plus sign. Putting the given number in the equation and using the values for the constants given in the textbook, we have
EB = (15.75 MeV)(86) – (17.80 MeV)(86)2/3 – (0.71 MeV)(36)(35)/861/3 – (23.69 MeV)(86 – 72)2/86 + (39 MeV)(86)-4/3
EB = 751.1 MeV and EB/A = 751.1 MeV/86 = 8.73 MeV/nucleus.
80Ta73: A = 180 and Z = 73. N = A – Z = 107, which is odd.
EB = (15.75 MeV)(180) – (17.80 MeV)(180)2/3 – (0.71 MeV)(73)(72)/1801/3 – (23.69 MeV)(180 – 146)2/86 + (39 MeV)(180)-4/3
EB = 1454.4 MeV and EB/A = 1454.4 MeV/180 = 8.08 MeV/nucleus.
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