468 8 Nuclear Physics – II
When the target and the projectile are interchanged, the same excitation
energyW∗produced withTais given by
−W∗=ETamp
mp+mTa(2)
Comparing (1) and (2)
ETa+EpmTa
mp= 897 .7MeV8.64 Eth=|Q|
(
1 +
mo
mca)
= 7. 83
(
1 +^1648
)
= 10 .44 MeVThe velocity,β=υ
c=
√
2 T
mc^2=
√
2 × 10. 44
16 × 931
= 3. 74 × 10 −^2
The impact parameter,b=R 1 +R 2 = 1. 1(
16
(^1) /
(^3) + 48
(^1) /
3
)
= 6 .153 fmJ=Moυb=nn=Moυb
=
Moc^2 βb
c=
16 × 931 × 3. 74 × 10 −^2 × 6. 153
197. 3
= 17 .37 or 178.3.11 Cross-sections ....................................
8.65 If 1% of neutrons are absorbed then 99% are transmitted. The transmitted
number I are related to the incident number by
I=I 0 exp(−μx)(1)
whereμis the absorption coefficient andxis the thickness of the foil.
μ=Σ=σN=σN 0 ρ/A
whereσis the microscopic cross-section,N 0 is the Avagardro’s number,ρthe
density andAthe atomic weight.
μ=28000 × 10 −^24 × 6. 02 × 1023 × 7. 3
114. 7
cm−^1= 1 ,073 cm−^1
I/I 0 = 99 / 100 =exp(− 1 , 073 x)
x= 9. 37 × 10 −^6 cm or 9. 37 μm8.66 Specific activity, that is activity per gram
|dQ/dt|=Qmaxλ=φΣact=φσaN 0 /A
= 5 × 1012 × 20 × 10 −^24 × 6. 02 × 1023 / 60
= 1012 disintegrations per second per gram.8.67|dQ/dt|=Qλ=φΣact(1−e−λt)
Given|dQ/dt|/|dQs/dt|= 20 / 100 = 1 −e−^0 .693t/^5.^3
or exp(0. 693 t/ 5 .3)= 1. 25
Take logeto findt= 1 .7 years.