Examples with hints
| Examples with hints for lectures 5 to 7 |
| m1t2.1 A heavy nucleus (Z, A) is symmetrically broken into two lighter nuclei. What is the expression of energy release in terms of B.E. equation? |
| Hint: Symmetrically broken means a nucleus of Z, A breaks into two nuclei with Z/2, A/2. Calculate energy release from the following equationErelease = B.E. (Z, A) – 2 B.E. (Z/2, A/2). Now use eqn. m1.27 and get an explicit expression for Erelease. |
| m1t2.2 Using Weizsacker’s semi empirical mass formula get an expression for the nuclear charge corresponding to most stable isobar for odd A nuclei. . |
| Hint: Express mass formula in the following manner M = aA + bZ +cZ2 /img1.gif) δThen use the condition (∂M/∂Z)constant A = 0 [Remember for odd A nuclei, δ = 0]You will get Zstable = -b/2c[Check the expression for, a, b and c] |
| m1t2.3 For odd A isobar series find M (Z,A) – M (Zstable, A). |
| Hint: From Zstable = -b/2c, you will findM (Zstable, A ) = aA –cZstable2Now try to get the relation: M (Z, A) – M (Zstable, A) = c (Z-Zstable)2 |
| m1t2.4 What is the mass difference when a β- particle is emitted from a nucleus in the above case? |
| Hint: The β – particle emission is nothing but the emission of electron from a nucleus and therefore in such process Z changes by one unit, i.e. Z+1.Now from M (Z, A) – M (Z+1, A), check whether you getMass difference = 2c (Zstable – Z – 1/2) |
| m1t2.5 Calculate binding energies of the following isobars and their binding energy per nucleon:28Ni64 = 63.9280; 29Cu64 = 63.9298; 30Zn64 = 63.9292.Assume, Mn = 1.009 amu and Mp = 1.008 amu |
| Hint: For 28Ni64 , N = 36 and Z = 28Now, NMn and ZMp = 36 x 1.009 + 28 x 1.008 = 36.324 + 28.224 = 64.548 amuTherefore, Δ M = 64.548 – 63.928 = 0.620 amu Hence B. E. = 0.620 x 931 MeV = 577.2 MeV And, B. E./A = 9.02 MeV Do the same for other two nuclei. Can you comment which nucleus is active for both beta particles. |