Nuclear Binding Energy

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Dwain Hart
5 years ago
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1 Nuclear Energy Nuclei contain Z number of protons and (A - Z) number of neutrons, with A the number of nucleons (mass number) Isotopes have a common Z and different A The masses of the nucleons and the electron are m p = kg = u = MeV m n = kg = u = MeV m e = kg = u = MeV

2 Nuclear Binding Energy The mass of the helium atom is u Nuclei heavier than hydrogen are created through nuclear fusion Conversely, the decay of heavier nuclei into lighter daughter nuclei is called fission The combined mass of two individual neutrons and two individual protons is m He = u The difference of u is called the binding energy and can be expressed as E b = m c 2 = MeV

3 Nuclear Timescale Binding energy is 0.7% of helium mass E b = MeV = E He Assume the Sun to be originally composed to 100% of hydrogen and that the inner 10% of it s mass converted into helium E Nuclear = M Sun c 2 = J For the nuclear timescale we get E Nuclear t Nuclaer = ~ yr L Sun

4 Strong Nuclear Force In the interaction of two positively charged protons the repulsive Coulomb interaction dominates at large distances beyond 1 fm At shorter distances than ~1 fm the strong nuclear force is attracting the two nucleons This force is responsible for the formation of stable nuclei

5 Quantum Mechanical Tunneling The thermal energy of hydrogen atoms even in the center of the Sun is not high enough to overcome the Coulomb barrier The Heisenberg Uncertainty Principle allows the possibility of a proton to be inside the central potential of another hydrogen nucleus Temperatures inside the Sun of ~10 7 K are high enough for the tunneling to occur Classically x p x ћ / Z 1 Z 2 e 2 k T = 4 π ε0 r T classical ~ K

6 Quantum Mechanical Tunneling Using one wavelength as the distance of closest approach where the Coulomb potential barrier is equal to the kinetic energy Z 1 Z 2 e 2 ( h / λ ) 2 = 4 π ε 0 λ 2 µ m Solving this equation for λ and using it for r in 3 2 Z 1 Z 2 e 2 k T = 4 π ε0 r T quantum = Z 1 2 Z 2 2 e 4 µ m 12 π 2 ε 02 h 2 k ~ 10 7 K with the reduced mass of the two protons µ m = m p / 2

7 Nuclear Reaction Rates Using the Maxwell-Boltzmann velocity distribution for the energy distribution of non-relativistic particles Kinetic energy describes the total energy accurately enough K = E = µ m v 2 2 Maxwell-Boltzmann distribution for number of particles within a unit volume with energies between E and E + de n E 2 n de = 1 3/2 1/2 E / k T E e de 1/2 π k T This equation does not describe the probability of how many of the particles will interact

8 Nuclear Reaction Rates Define the (nuclear reaction) cross-section as σ (E ) = Number of reactions / nucleus / time Number of incident particles / area / time Number of incident particles per unit volume with energies between E and E + de is given by n ie de Number of reactions dn E is the number of particles that interact in a time interval dt with a velocity v ( E ) dn E = σ (E ) v(e ) n ie de dt

9 Nuclear Reaction Rates Number of incident particles per unit volume with appropriate kinetic energy is with n i = 0 n ie de = n ie de n i n and n E de n = 0 n E de Results in dn E = σ (E ) v(e ) n n i n E de dt

10 Nuclear Reaction Rates Number of reactions per target nucleus per time interval dt is dn E dt n = σ (E ) v(e ) i n E de n If there are n x target nuclei per unit volume, then the total number of reactions per unit volume per unit time, integrated over all energies is r ix = n x n i σ (E ) v(e ) de n 0 To evaluate this integral, we need to know the cross-section, which varies rapidly with energy and has a fairly complicated functional form n E

11 Nuclear Reaction Rates The cross-section can be regarded as a physical area The size of the nucleus in terms of interactions is roughly the de Broglie wavelength r ~ λ σ (E ) π λ 2 π ( h / p ) 2 1 E The tunneling probability depends on the Coulomb barrier height U c σ (E ) e 2 π 2 U c / E σ (E ) e b E 1/2 with b = µ m 2 1/2 π Z 1 Z 2 e 2 ε 0 h

12 Nuclear Reaction Rates Combining the two above results and defining a function S(E) containing the remaining slow energy variation S (E) σ (E ) = e b E 1/2 E Using this we obtain for the reaction rate integral r ix = 2 3/2 n x n i b E S (E ) e 1/2 E / k T e de 1/2 k T ( µ m π ) 0 Coulomb barrier penetration Maxwell-Boltzmann

13 Gamov Peak r ix = 2 3/2 n x n i b E S (E ) e 1/2 E / k T e de 1/2 k T ( µ m π ) 0 The maximum occurs at E 0 = b k T 2 2/3 which is called the Gamov Peak The nuclear reaction has its maximum contribution in a relatively small energy range

14 Energy Generation Rate Assume E 0 to be the amount of energy released per reaction, then the amount of energy liberated per kilogram of material per second is ε ix = E 0 ρ r ix The luminosity of a star is given by total energy E released per kilogram and per second by all nuclear reactions and gravity dl = ε dm = ε 4 π r 2 ρ dr ε = Σ ε ix dl r dr = 4 π r 2 ρ ε Describes the change of the interior luminosity L r due to total energy generated

[2] State in what form the energy is released in such a reaction.... [1]

[2] State in what form the energy is released in such a reaction.... [1] (a) The following nuclear reaction occurs when a slow-moving neutron is absorbed by an isotope of uranium-35. 0n + 35 9 U 4 56 Ba + 9 36Kr + 3 0 n Explain how this reaction is able to produce energy....