1000 Solved Problems in Modern Physics

422 7 Nuclear Physics – I

Usingλ 238 = 1 /τ 238 = 1 / 6. 52 × 109 ,λ 235 = 1 /τ 235 = 1 / 1. 02 × 109

And solving (3) we get the age of the earth as 5. 26 × 109 years

7.99 Number of Th atoms in 2μgis

N=N 0 W/A= 6. 02 × 1023 × 2 × 10 −^6 / 224 = 5. 375 × 1015

λ= 0. 693 /(3. 64 × 86 ,400)= 2. 2 × 10 −^6

ActivityA=|dN/dλ|=Nλ=(5. 375 × 1015 )(2. 2 × 10 −^6 )= 1. 183 ×

1010 /s

=(1. 188 × 1010 / 3. 7 × 1010 )ci= 0 .319 ci

7.100 1 rad=100 ergs g−^1

Energy received=(40)(100)(20× 103 )/ 107 J

=8J

= 8 / 4. 18 = 1 .91 Cal

7.101 LetVcm^3 be the volume of blood, Initial activity is 16, 000 /Vper minute

per cm^3 .0. 8 =^16 ,V^000 exp(− 0. 693 × 30 /15)

=^4 ,^123 V

∴V=5000 cm^3

7.3.10 Alpha-Decay ....................................

7.102λ 1 = 1021 exp(− 2 πzZ 1 e^2 /v 1 )

λ 2 =^1021 exp(−^2 πzZ 2 e^2 /v 2 )

Givenλ 1 =λ 2

Z 1 /

√

E 1 =Z 2 /

√

E 2

E 2 =E 1 Z 22 /Z 12 = 5 .3(80/82)^2 = 5 .04 MeV

Thus energy from the second nuclei is 5.04 MeV.

7.103 The energy required to force anα-particle(classically) into a nucleus of
chargeZeis equal to the potential energy at the barrier height and is given by
E=zZe^2 / 4 πε 0 R.
This is barely possible whenαparticle and the Uranium nucleus will just
touch each other and the kinetic energy of the bombarding particle is entirely
converted into potential energy.
R=R 1 +R 2 =r(A
1 / 3
1 +A^2

1 / (^3) )
= 1 .2(4^1 /^3 + 2381 /^3 )= 9 .34 fm
E= 1. 44 zZ/R= 1. 44 × 2 × 92 / 9. 34 = 28 .37 MeV
7.104 Geiger–Nuttal law is
logλ=klogx+c
wherekandcare constants,λis the decay constant andxis the range
log(0. 693 /T)=klogx+c
k logx+logT=log 0. 693 −c=c 1 (1)