Bharat Ki Samasyae Evam Samadhan
- Date Submitted: 01/12/2012 09:14 PM
- Flesch-Kincaid Score: 105.1
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THE NUCLEUS
CHAPTER - 46
1. M = Amp, f = M/V, mp = 1.007276 u 1/3 –15 1/3 –27 R = R0A = 1.1 10 A , u = 1.6605402 10 kg
4 / 3 3.14 R3 14 ‘f’ in CGS = Specific gravity = 3 10 .
2. f=
=
A 1.007276 1.6605402 10 27
= 0.300159 10
18
= 3 10
17
kg/m .
3
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8. 9.
M M 4 1030 1 1 V 1013 1014 17 v f 0.6 6 2.4 10 3 V = 4/3 R . 1 1 3 1 3 3 1014 = 4/3 R R = 1014 6 6 4 1 100 3 R = 1012 8 4 4 R = ½ 10 3.17 = 1.585 10 m = 15 km. Let the mass of ‘’ particle be xu. ‘’ particle contains 2 protons and 2 neutrons. 2 Binding energy = (2 1.007825 u 1 1.00866 u – xu)C = 28.2 MeV (given). x = 4.0016 u. 7 7 Li + p l + + E ; Li = 7.016u 4 = He = 4.0026u ; p = 1.007276 u 7 E = Li + P – 2 = (7.016 + 1.007276)u – (2 4.0026)u = 0.018076 u. 0.018076 931 = 16.828 = 16.83 MeV. 2 B = (Zmp + Nmn – M)C Z = 79 ; N = 118 ; mp = 1.007276u ; M = 196.96 u ; mn = 1.008665u 2 B = [(79 1.007276 + 118 1.008665)u – Mu]c = 198.597274 931 – 196.96 931 = 1524.302094 so, Binding Energy per nucleon = 1524.3 / 197 = 7.737. 238 4 234 a) U 2He + Th E = [Mu – (NHC + MTh)]u = 238.0508 – (234.04363 + 4.00260)]u = 4.25487 Mev = 4.255 Mev. 238 234 b) E = U – [Th + 2n0 + 2p1] = {238.0508 – [234.64363 + 2(1.008665) + 2(1.007276)]}u = 0.024712u = 23.0068 = 23.007 MeV. 223 209 14 Ra = 223.018 u ; Pb = 208.981 u ; C = 14.003 u. 223 209 14 Ra Pb + C 223 209 14 m = mass Ra – mass ( Pb + C) = 223.018 – (208.981 + 14.003) = 0.034. Energy = M u = 0.034 931 = 31.65 Me. 1 EZ.N. EZ–1, N + P1 EZ.N. EZ–1, N + 1H [As hydrogen has no neutrons but protons only] 2 E = (MZ–1, N + NH – MZ,N)c
E2N = EZ,N–1 + 1 n . 0 Energy released = (Initial Mass of nucleus – Final mass of nucleus)c = (MZ.N–1 + M0 – MZN)c .
2 2
10. P32 S32 +
0v
0
10
Energy of antineutrino and -particle = (31.974 – 31.972)u = 0.002 u = 0.002 931 = 1.862 MeV = 1.86. –...
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