Words of Wisdom:

"It's not how smart you are, but how hard you try" - Teacher

Bharat Ki Samasyae Evam Samadhan

  • Date Submitted: 01/12/2012 09:14 PM
  • Flesch-Kincaid Score: 105.1 
  • Words: 3590
  • Essay Grade: no grades
  • Report this Essay
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

3.

4.

5.

6.

7.

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 + 2n0 + 2p1] = {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

 10

Energy of antineutrino and -particle = (31.974 – 31.972)u = 0.002 u = 0.002  931 = 1.862 MeV = 1.86. –...

Comments

Express your owns thoughts and ideas on this essay by writing a grade and/or critique.

  1. No comments