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 1026 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
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 1026 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