Nuclear_Physics_Lecture_7.pdf

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(1)α – Decay: Explanation The α – particle faces Coulomb potential barrier surrounding the nucleus For. explanation,. we. assume. rectangular barrier The average barrier height R. 2 Ze 2 / r V =∫ dr b−R b The width of the barrier a = (b – R) Transmission probability 16 E (V0 − E ) T= exp( −2αa ), 2 V0. 2 M (V0 − E ) α= h. a.

(2) α – Decay: Explanation A typical example: α – emission from 238U R = 8.7 × 10-13 cm, b = 61.7 × 10-13 cm ⇒ V = 9.86 MeV E = 4.2 MeV, a = 5.3 × 10-12 cm. Calculation based on these data yields too low a probability of penetration through the barrier compared to the experimental value Instead, we take typical values V0 = 16 MeV, E = 6 MeV, a = 2 × 10-12 cm ⇒ αa =. 2 M (V0 − E ) .a h. [ 2 × 4 × 1.66 × 10 =. − 27. × (16 − 6) × 1.6 × 10−13 1.05 × 10−34. ]. 1/ 2. × 2 × 10 −14 = 27.76.

(3) α – Decay: Explanation 16 E (V0 − E ) exp( −2αa ) T= 2 V0 =. 16 × 6 × (16 − 6) exp( −2 × 27.76) = 3 × 10−24 Still too low (apparently)! 16 × 16. Not like that! Number of attempts made per second n = velocity/nuclear diameter. 1.7 × 107 20 = = 8 . 5 × 10 2 × 10−14 20 −24 −3 Probability of escape per second P = nT = 8.5 × 10 × 3 × 10 = 2.55 × 10. Mean lifetime against the α – decay 1 1 τ= = sec = 6 min 32 sec −3 P 2.55 × 10 Pretty acceptable figure!.

(4) β – Decay β-decay is a type of radioactive decay in which an energetic electron or positron, and a neutrino are emitted from a nucleus. For example, beta decay of a neutron transforms it into a proton by the emission of an electron, or conversely a proton is converted into a neutron by emission of a positron, thus changing the nuclide type. Neither the beta particle nor its associated neutrino exist within the nucleus prior to beta decay, but are created in the decay process. By this process, unstable atoms obtain a more stable N/Z ratio. The probability of a nuclide decaying due to beta decay is determined by its binding energy..

(5) β – Decay β– – decay A neutron is converted to a proton and the process creates an electron and an electron antineutrino inside the nucleus Since the mass number and atomic number are conserved, the atomic number of parent nucleus increases from Z to Z+1 & neutron number decreases from N to N – 1. n → p + e− + υe − A A X → Y + e + υe Z Z +1. Examples:. 14 14 − C → N + e + υe 6 7 137 137 − Cs → Ba + e + υe 55 56.

(6) β – Decay β– – decay: Disintegration energy − A A X → Y + e + υe Z Z +1 Q value of this reaction is. Qβ − = [M n ( A, Z ) − M n ( A, Z + 1) − me ]c 2 Mass of neutrino in ignored. Instead of nuclear masses, if we take atomic masses and those expressed in energy unit (MeV). Qβ − = M ( A, Z ) − Zme − M ( A, Z + 1) + ( Z + 1)me − me. = M ( A, Z ) − M ( A, Z + 1) For spontaneous β– – emission. Qβ − > 0. ⇒ M ( A, Z ) > M ( A, Z + 1).

(7) β – Decay β+ – decay A proton is converted to a neutron and the process creates a positron and an electron neutrino Since the mass number and atomic number are conserved, the atomic number of parent nucleus decreases from Z to Z – 1 & neutron number increases from N to N + 1. p → n + e+ + υe + A A X → Y + e + υe Z Z −1. Examples:. 23 23 + Mg → Na + e + υe 12 11 80 80 + Br → Se + e + υe 35 34.

(8) β – Decay β+ – decay: Disintegration energy + A A X → Y + e + υe Z Z −1 Q value of this reaction is. Qβ + = [M n ( A, Z ) − M n ( A, Z − 1) − me ]c 2 Mass of neutrino in ignored. Instead of nuclear masses, if we take atomic masses and those expressed in energy unit (MeV). Qβ + = M ( A, Z ) − Zme − M ( A, Z − 1) + ( Z − 1)me − me. = M ( A, Z ) − M ( A, Z − 1) − 2me For spontaneous β+ – emission Qβ + > 0. ⇒ M ( A, Z ) − M ( A, Z + 1) > 2me (1.022 MeV ).

(9) β – Decay Decay of a free neutron into a proton is allowed but decay of free proton into a neutron is not allowed.. n → p + e− + υe. √. p → n + e+ + υe Both the reactions are allowed by charge conservation, spin angular momentum conservation and conservation of other fundamental quantum numbers (like lepton number, baryon number) However, the rest mass of neutron is higher than the rest mass of proton. Hence from the conservation of mass-energy principle, the 1st reaction is allowed but the 2nd reaction is forbidden.

(10) β – Decay Orbital electron capture A process in which a nucleus captures one of its atomic electrons, resulting in the emission of a neutrino The atomic number of parent nucleus decreases from Z to Z – 1 & neutron number increases from N to N + 1 A − A X + e → Y + υe Z Z −1. Examples:. 7 − 7 Be + e → Li + υe 4 3 80 − 80 Br + e → Se + υe 35 34.

(11) β – Decay Electron capture: Disintegration energy A. Z. −. A. X + e → Z −1Y + υ e. Q value of this reaction is. Binding energy of the orbital electron – few tens of eV. Qe = [M n ( A, Z ) + me − M n ( A, Z − 1)]c 2 − Be Mass of neutrino in ignored. Instead of nuclear masses, if we take atomic masses and those expressed in energy unit (MeV). Qe = M ( A, Z ) − Zme + me − M ( A, Z − 1) + ( Z − 1)me − Be. = M ( A, Z ) − M ( A, Z − 1) − Be For spontaneous capture process Qe > 0. ⇒ M ( A, Z ) − M ( A, Z − 1) > Be.

(12) β – Decay A nucleus energetically allowed to undergo β – decay can undergo orbital electron capture, but the not the vice – versa In both processes (β+ – decay & electron capture) atomic number decreases by one unit. In β+ – decay the difference between the atomic masses of parent and daughter nuclei must exceed twice the mass of electron which is 1.022 MeV.. M ( A, Z ) − M ( A, Z + 1) > 2me (1.022 MeV ) However, in orbital electron capture the difference between the atomic masses of parent and daughter nuclei must exceed the binding energy of the orbital electron (Be) which is only few tens of eV.. M ( A, Z ) − M ( A, Z − 1) > Be.

(13) β – Decay Problem: 7 7 4Be and 3Li have atomic masses 7.01693 u and 7.01600 u respectively. Which of them shows −activity and of what type? Justify your answer. Given, M(Be7) = 7.01693 u & M(Li7) = 7.01600 u [CU – 2013] Li7 can not undergo β– – decay to transform into Be7 since M(Li7) < M(Be7) Be7 can undergo β – decay to transform into Li7 via two possible ways: (i) β+ – decay & (ii) orbital electron capture For spontaneous β+–decay the condition is M(Be7)–M(Li7) > 2me (1.022 MeV) For spontaneous orbital electron capture, the condition is M(Be7)–M(Li7) > Be (the binding energy of the orbital electron which is few tens of eV) Now, M(Be7) – M(Li7) = (7.01693 – 7.01600) u = 0.00093 u = 0.866 MeV Thus transformation through β+ – decay is not energetically allowed. Be7 can undergo orbital electron capture to transform into Li7.

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