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# Lamarsh Solution The whole doc is available only for registered users
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• Category: Atomic

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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

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



2.27

There are “a ground state at the buttom and three excited states “following it.And the possible gamma rays are, E1=0.403-0.208=0.195 MeV E4=0.403-0.158=0.245 MeV E2=0.208-0.158=0.005 MeV E5=0.208-0=0.208 MeV E3=0.158-0=0.158 MeV E6=0.403-0=0.403 MeV

2.29 (a)
3 1

H     23He  

(b)



ln 2  1.79e  9sec onds 1 after that we know at any time t the activity is, 12.26 years, the activity of Tritium in question is ,1mCi=3.7e7 dis/sec ;by using the activity formula we can find the # of Tritium atoms,

3.7e7  2.06e16 1.79e  9

atoms, and the mole # of tritium and then the mass of it found as,  # of moles= 2.33

2.06e16 =3.43e-8 and finally mass=3.43e-8 x 3.016 grams=1.03e-7gr 0.6022e24

We’ll again use the relation

   N . Here using half-lifes,
and finally using above formula we find the # of atoms per cm3 as,

Cs 137  7.27e  10 Cs 134  1.06e  8
NCs 137  NCs 134

156e  6 Ci x 3.7e10  7.93e15 atoms/cm3 7.27e  10 26e  6 Ci x 3.7e10   9.07e13 atoms/cm3 1.06e  8

2.37 We know at time t,which is asked,that

N 25  0.03 and now this ratio is, N 25  N 28

N  0.0072 , N  N 28*
* 25

* 25

N 25*  N 25e 25t
in here

N 28*  N 28e 28t

And using these relations after some maths manipulation we found that,

7.25e  3N 28* N 25  e( 28t 25t ) * N 28 32.33N 25 2.40

from here the time is,1.6e9(billion)years ago

dN A =-λ A N A dt dN B =λ A N A dt-λ B N Bdt from here N A (t)=N 0 e-λA t and N B (t)=N 0 λA (e-λA t -e-λB t ) λB  λA

As mentioned in the question , λ A

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