424 7 Nuclear Physics – I
Dividing (3) by (2)
N 2 /N 1 =[1−exp(−λt 2 )]/[1−exp(−λt 1 )]
N 2 /N 1 = 2. 66 =[1−exp(− 6 λ)]/[1−exp(− 2 λ)]
=(1−x^3 )/(1−x)= 1 +x+x^2 (4)
wherex=exp(− 2 λ)
Solving the quadratic equation forx, we findx= 0. 885 =exp(− 2 λ)
whence the mean life timeτ= 1 /λ= 15 .93 s7.108 According to Fermi’s theory ofβ-decay, forE 0 >>mec^2 , the decay constant
λ=G^2 |Mif|^2 E^50
60 π^3 (c)^6 (1)
So that with the valueG/^3 c^3 = 1. 166 × 10 −^5 GeV−^2 , (1) becomesλ=
1. 11 E^50 |Mif|^2
104 S− (^1) =^1.^11 ×(3.5)^5 ×^6
104
= 0 .3466 s−^1
Therefore,T 1 / 2 = 0. 693 / 0. 3466 = 2 .0 s. The experimental value is 0.8 s.
7.109 The mean energy of electrons
∫Emax
0
E n(E)dE/
∫ Emax
0
n(E)dE
Givenn(E)dE=c
√
E(Emax−E)^2 dE
wherec=constant<E>=c∫Emax
0 E√
E(Emax−E)^2 dE
c∫Emax
0√
E(Emax−E)^2 dE=
Emax
3
If all the electrons emitted are absorbed then the kinetic energy of the
electrons is converted into heat.
Heat evolved/sec = (mean energy)(no. of electrons emitted/second)
=(0.156)×(3. 7 × 107 )/3MeV/s= 1. 92 × 106 MeV/s7.110 Let the threshold energy beEν. In that case the particles in the CMS will be
at rest. Now, in the neutron decay
n→p+e−+ν ̄+ 0 .79 MeV
Mn−(Mp+Me)= 0. 79
as ̄νhas zero rest mass
Mn+Me=Mp+ 2 me+ 0. 79 =Mp+ 1. 81
(the masses of electron and positron are identical, each equal to 0.511 MeV)
We use the invariance of
(ΣE)^2 −|ΣP|^2 =(ΣE∗)^2 −zero
where (∗) refers to the CMS and total momentum of particles in CMS is zero
(Ev+Mp)^2 =(Mn+Me)^2 =(Mp+ 1 .81)^2
OrEν= 1 .81 MeV.