Lamarsh Solutions

Chapter-3 Part-2

3.26

(a) ' 2 2 2

2[cos sin ] 0.846

( 1)

EE A Mev

A

(b) ' 0.154AE E E Mev

(c) conservation of momentum:

' '

' '

cos cos

0 sin sin

mv mv MV

mv MV

Cons. of energy: 2 '2 '20.5 0.5 0.5mv mv MV and finally,

01arccos( 0.5( 1)) 42.61A

3.30

(a)

2( 1) 11 ln( )

2 1

A A

A A

and 1H and knowing that,

=

2 6ln( )

U 1 14.51

e ev

ev collisions

(b)

0.158C and =

2 6ln( )

U 1 91.820.158

e ev

ev collisions

3.39

5.4s e i b 6.7t s a b

0.1a f 0.0833f

3.49

(a)Actually in here we are asked the burnup rate from Lamarsh eq. 3.57

Burnup rate=1.05P g/day=1.05 x 2758=2906.4 g/day

(b) . . . . (1 )f ca

f f

cons rate fis ratex fis ratex fis ratex

Cons.rate=1.05P x (1+0.169)=3397.6 gr-U235/day

3.58

2 3

0 0

0

1.84 / 11.34 /

( ) or ( )

I ln(1000)and then ln( ) this gives that 0.331

I(x) 1.84 11.34

Pb

xx

cm gr gr cm

I x I e I x I e

x x cmx

3.61

Cross sections are given in terms of per electron;so we can find both for Al and water the total

cross sections as and e e

c c ca caxZ xZ and using these cross sections well find the total

cross sections or attenuation coefficients as Nx .And finally using the appropriate

coefficients and W ExIx ,will find the deposited energies.Notice that well not use the total compton cross section for this calculation ,0.4929b, but use the compton abs. cross section for

energy absorbtion due to compton ; for energy abs. due to photelectric effect well use the total

mass attenuation coefficient at energy 0.1 Mev to find the total photoel. absorption coefficient

(a) Compton scattering.

(i)For Al, 2 20.0685 24 13 0.8905 24ca e cm x e cm and

3 2 10.06024 24 / 0.8905 24 0.05363ca e atoms cm x e cm cm

30.1 5 6 0.05363 4.2949 9

WW ExIx Mevx e x e

cm

(ii)For water, 2 20.0685 24 (8 2) 0.685 24 andca e cm x e cm

3 2 10.03342 24 / 0.685 24 0.0228ca e atoms cm x e cm cm

30.1 5 6 0.0228 1.834 9

WW ExIx Mevx e x e

cm

(b)Photoelectric Effect,

(i)For Al,

1

,

1

, , ,

( ) @0.1 0.0373 2.699 0.1006

0.1006 0.0536 4.707 2

Al tot a

Al photoel Al tot Al compton

Mevx x cm

e cm

And finally deposited energy is,

30.1 5 6 4.707 2 3.7703 9

WW ExIx Mevx e x e e

cm

(ii)For water,

1

,

1 1 1

, , ,

( ) @0.1 2.53 2

2.53 2 2.2897 2 2.403 3

water tot a

water photoel water tot water compton

Mevx e cm

e cm e cm e cm

And finally deposited energy is,

1

30.1 5 6 2.403 3 1.9248 10

WW ExIx Mevx e x e cm e

cm

WARNING!!!!!:In energy calculation ,you finally found the unit in Mev ,so you must convert it

into joule to find Watt.In here this is left to the student!!!