1000 Solved Problems in Modern Physics

7.3 Solutions 403

7.30 The fraction of particles scattered in brass at angles exceedingθ is given
byΔN/N =(π/4)(1. 442 /T^2 )

(

0. 7 Z 12 /A 1 + 0. 3 Z 22 /A 2

)

ρtN 0 cot^2 θ/ 2 ×

10 −^26 whereZ 1 =29 for copper andZ 2 =30 for Zinc,A 1 = 63 .55 and

A 2 = 65 .38 are the atomic masses, respectively, andN 0 = 6. 02 × 1023 is the

avagadro’s number,T= 1 .5MeV,ρt= 2 × 10 −^3 gcm−^2 andθ= 450 .The

factor 10−^26 is introduced to convert fm^2 into cm^2. Using these values in the

above equationΔN/N= 6. 78 × 10 −^4

7.31 (a)Δσ=σ(90◦, 180 ◦)=(π/4)(1. 44 zZ/T)^2
UsingΔσ= 0 .6kb=60 fm^2 ,z= 2 , Z=79 and solving forT, we get
T= 3 .36 MeV
(b)σ(θ)=(1/16)(1. 44 zZ/T)^21 /sin^4 (θ/2)
Putθ = 90 ◦ andT = 3 .36 MeV to findσ(90) = 1 ,146 fm^2 /sr =
11 .46 b/sr.

7.32 First we work out in the CMS and then transformσ(θ∗)intheCMStoσ(θ)
in the LS.
Rutherford’s formula in CMS is
σ(θ∗)=(1/4)(zZe^2 /μv^2 )^2. 1 /sin^4 (θ/2) (1)

where the reduced mass

μ=mM/(m+M)=m/(1+γ)(2)

wheremandMare the masses of the incident and target masses, respectively,

andγ=m/M.

Further

sin^4 (θ∗/2)=(1/4)(sin^4 θ∗)/(1+cosθ∗)^2 (3)

and

tanθ=sinθ∗/(γ+cosθ∗)(4)

Squaring (4) and expressing it as a quadratic equation and solving it

cosθ=γ sin^2 θ±cosθ(1−γ^2 sin^2 θ)^1 /^2 (5)

Nowσ(θ)=(1+γ^2 + 2 γcosθ∗)^3 /^2 /| 1 +γcosθ∗| (6)

Combining (1), (2), (3), (5) and (6) and after some algebraic manipulations

we get

σ(θ)=

(

zZe^2

2 T

) 2

1

sin^4 θ

[cosθ±(1−γ^2 sin^2 θ)^1 /^2 ]^2

(1−γ^2 sin^2 θ)^1 /^2

Ifγ<1, the positive sign only should be used before the square root. If

γ>1, the expression should be calculated for positive and negative signs and

the results added to obtainσ(θ). Forγ= 1 ,σ(θ)=

(

zZe^2

T

) 2

cosθ

sin^4 θ