Nuclear_Physics_Lecture_6.pdf

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(1)Unstable Nuclei Syllabus Alpha decay: alpha particle spectra – velocity and energy of alpha particles. Geiger-Nuttal law Beta decay: nature of beta ray spectra, the neutrino, energy levels and decay schemes, positron emission and electron capture, selection rules, beta absorption and range of beta particles, Kurie plot Gamma decay: gamma ray spectra and nuclear energy levels, isomeric states. Gamma absorption in matter – photoelectric process, Compton scattering, pair production (qualitative).

(2) α – Decay Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle (helium nucleus) and thereby transforms or 'decays' into an atom with a mass number that is reduced by four and an atomic number that is reduced by two. An alpha particle is identical to the nucleus of a helium-4 atom, which consists of two protons and two neutrons. It has a charge of +2e and a mass of ~4 u. A A− 4 4 X → Y + He Z Z −2 2.

(3) α – Decay Like He4 nucleus, the α – particles have their spin 0 and hence follow BE statistics Alpha particles have their initial kinetic energy of the order of few MeV (typically 3 MeV – 7 MeV) and velocity of the order of 107 m/s.

(4) α – Decay Disintegration Energy. A A− 4 4 X → Y + He Z Z −2 2. Qα = M X ( A, Z ) − M Y ( A − 4, Z − 1) − M α ( A = 4, Z = 2) Linear momentum conservation. M YVY + M αVα = M XVX = 0 Mα ⇒ VY = − Vα MY 2.  Mα  1 1 1 1 2 2 Qα = M YVY + M αVα = M Y  − Vα  + M αVα2 2 2 2  MY  2  Mα  1 Mα  2  = Eα 1 +  = M αVα 1 + 2  MY   MY  Qα MY  A−4 ⇒ Eα = = Qα ⋅ ≈ Qα ⋅   Mα MY + Mα  A  1+ MY.

(5) α – Decay Problem: Polonium – 212 emits α particles whose kinetic energy is 10.54 MeV. Determine the α – disintegration energy. [CU – 2017]. MY  A−4 Eα = Qα ⋅ ≈ Qα ⋅   MY + Mα A    212   A  ⇒ Qα = Eα  = 10 . 54 ×    MeV ≈ 10.74 MeV  212 − 4   A−4.

(6) α – Decay Fine Structure of α –Spectrum A− 4. ∗ 4 X → Y + He 2 Z Z −2 A. A− 4. ∗. A− 4. Y → Z − 2Y + γ. Z −2. The daughter nucleus is produced in different excited states which subsequently comes down to ground state by releasing γ - photons.

(7) α – Decay Range In passing through matter, the α – particles ionize and thus lose energy in many steps, until their energy reduces to (almost) zero. The distance to this point is called the range of the particle (R) The range depends on (i) Initial energy of the α – particles (ii) Ionization potential of the medium through which it passes (iii) Collision probability of α – particles with the atoms/molecules of the medium (controlled by pressure, temperature, density etc. of the medium).

(8) α – Decay Range R is measured in the unit of length – cm or m For α – particles with kinetic energy few MeV, R is few mm in solids and liquids, and few cm in gas R can also be expressed in the unit of mass of the substance traversed (g/cm2 or kg/m2).

(9) α – Decay Range The amount of mass contained within a cylinder of unit cross-sectional area and length equal to the range of the α – particles is = R×ρ, ρ = density of the material Alternatively, R is measured in the unit of mass per unit area (g/cm2 or kg/m2).

(10) Measurement. Range. B corresponds to mean range C corresponds to extrapolated range CD – straggling of range: It arises due to statistical nature of interaction – number of collisions required to bring a radiation particle to rest within the medium will vary slightly for each. 210Po. α – rays (Eα = 5.3007 MeV). particle (i.e., some may travel further as. Mean range = 3.842 cm (air). undergo less collisions than others). Extrapolated range = 3.897 cm.

(11) Range vs. Energy: Geiger Law.

(12) Range vs. Energy: Geiger Law An empirical relationship between the range and energy of the α – particles R = aE 3 / 2 Since v ∝ E , R = bv 3. a = 0.315, b = 9.416 × 10–28 for E in MeV, R in cm. In solid the range Rs (in cm) is related to the range R (in cm) in air Rs =. 0.312 RA1 / 2. ρ. ρ = density of the solid A = mass number.

(13) Geiger Law Specific Ionization The number of ion pairs formed per unit path length at any point in the path of the α – particles is called specific ionization (I) dE dR Now, R = bE 3 / 2 or, E ∝ R 2 / 3 I∝. dE dE −1 / 3 ⇒ ∝ R or ∝ 1 / v (Q R ∝ v 3 ) dR dR. Specific ionization at a point is inversely proportional to the velocity of the α – particles at that point.

(14) Geiger Law Problem Assuming the specific ionization at point is inversely proportional to the velocity of the α – particles at that point, show that their range varies proportionally with third power of their velocity. dE 1 dE k ∝ ⇒ = − (k = constant) dx v dx v dE d  1 2  dv =  mv  = mv dx dx  2 dx  dv k ⇒ mv =− dx v at x = R, v = 0. On integration −. 0. R. m ∫ v 2dv = −k ∫ dx v. ⇒R=. 0. m 3 v = av 3 ( a = m / 3k = constant) 3k.

(15) Geiger – Nuttal Law An empirical relationship between the range (R) of the α –particles and the disintegration constant (λ) of the naturally α –active substances were established by Geiger & Nuttal in 1911 A, B = constants ln λ = A + B ln R ln 2 0 .693 λ= = T1 / 2 T1 / 2 T1/2 = half-life In terms of energy (since R = a E3/2) ln λ = C + D ln E.

(16) Geiger – Nuttal Law Implications It relates the range (R) with the disintegration constant or the halflife of a radioactive element It shows R is higher for higher λ (or smaller T1/2) i.e. for short-lived isotopes emit more energetic alpha particles than long-lived ones The relationship also shows that half-lives are exponentially dependent on decay energy, so that very large changes in half-life make comparatively small differences in decay energy, and thus alpha particle energy.

(17) Geiger – Nuttal Law Implications In practice, this means that alpha particles from all alpha-emitting isotopes across many orders of magnitude of difference in half-life, all nevertheless have about the same decay energy R can be measured directly. Hence, by measuring R, one can determine the disintegration constant or the half-life.

(18) Geiger – Nuttal Law Problem: Given that the range in standard air of the α – particles from radium (half-life = 1622 years) is 3.36 cm, whereas from polonium (half-life = 138 days) this range is 3.85 cm, Calculate the half-life of RaC’ for which the α – particle range is 6.97 cm. [CU – 2017]. ln 2 0.693 ln λ = A + B ln R, λ = = ⇒ ln 0.693 − ln T1/ 2 = A + B ln R T1/ 2 T1/ 2 or, ln T1/ 2 + B ln R = ln 0.693 − A = C (constant ) ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ (1) R = 3.36 cm for T1/2 = 1622 years = 1622×365 days = 592030 days & R = 3.85 cm for T1/2 = 138 days ⇒ ln 592030 + B ln 3.36 = C ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ (2). & ln138 + B ln 3.85 = C ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ (3) Solving equations (2) & (3), B = 61.46 unit, C = 38.12 unit Hence, from equation (1), for R = 6.97 cm ln T1 / 2 + 61.46 × ln 6.97 = 38.12. ⇒ ln T1 / 2 = −13.705 ⇒ T1 / 2 = 1.97 × 10−14 days = 1.7 ns.

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