Lamarsh Solution Chapter 2

Lamarsh Solutions Chapter-2 2.5 This is a question of probability, For molecules which have an approximate weight of 2, there are two 1H and we can find the probability or the percentage over 1 as, 0.99985*0.99985=0.99970 The same calculation can be made for the mol. weights of 3 and 4 For 3 there are one 1H and one 2 H and so, 0.99985*0.00015=1.49e-4 For 4 there are two 2 H and so, 0.00015*0.00015=2.25e-8 2.7 From table of nuclides we can find the atomic weights of O and H using the abundances

1 x[99.759 x15.99492  0.097 x16.99913  0.204 x17.99916] 100 m(O)  15.99938 1 m( H )  x[99.985 x1.007825  0.015 x 2.0141] 100 m( H )  1.007975 m( H 2O)  18.01533 m(O) 
(a) # of moles of water= (b) # of 1H atoms=2.76 moles x 0.6022e24 x 2 atoms of H x 0.99985 abund. =3.32e24 atoms of 1H (c) # of 2 H atoms=2.76 moles x 0.6022e24 x 2 atoms of H x 0.015e-2 abund. =4.98e20 atoms of 2 H

50 =2.76 moles 18.01533

2.16 The fission of the nucleus of
235

U releases approximately 200MeV. How much energy (in kilowatt235

hours and megawatt-days) is released when 1 g of Solution:

U undergoes fission?

1g 235 U 

1g  0.6022 1024 fissioned 235 Uatoms  2.563 1021 atoms 235 g
2.563 1021 atoms  200MeV  512.6 1021 MeV 1atoms

Released Energy:

 512.6 1021 MeV 
 22810kWhr 
2.20

4.450 1026 kWhr  22810kWhr 1MeV

1day  0.9505MWd 24hr

Erest  m0 c 2 Etot  mc 2 m m0 1

2 c2 ,in question we see the square of E_rest and E_tot ,so we do

Erest 2 m0c 2 2 ( ) ( 2 ) Etot mc

m0 2 Erest 2 ( ) ( ) 2 and from here we find m Etot m0 2 2 ( )  1 2 m c we find,

and knowing the

relation between m_0 and m as

Erest 2 1 2  c Etot 2
2 2

2

and finally

Erest 2   c (1  ) Etot 2
2.22

and

Erest 2   c 1 Etot 2

as wanted.

(a) wavelength of a 1 MeV photon can be found as,



1.24e  6 1.24e  6   1.24e  12m  1240 fm E 1e  6 eV
2.86e  9 2.86e  9   2.86e  12 cm  28.6 fm E 1e  6 eV