Physics and Engineering of Radiation Detection

2.4. Interaction of Heavy Charged Particles with Matter 109

scattered at 10^0 and 30^0.

dσ

dΩ

∣

∣

∣

θ=10^0

=3. 597 × 10 −^28

1

sin^4 (5^0 )

=6. 234 × 10 −^24 m^2 /ster

dσ

dΩ

∣

∣

∣

θ=30^0

=3. 597 × 10 −^28

1

sin^4 (15^0 )

=8. 016 × 10 −^26 m^2 /ster

Now that we have the differential scattering cross sections at the two angles,

we can multiply them by the solid angles subtended by the detector, which is

given by

Ω=

A

r^2

=

1

152

=4. 44 × 10 −^3 ster

to obtain the total cross sections for scattering at 10^0 and 30^0 as follows.

σ(θ)=

dσ

dΩ

Ω

⇒

σ(θ=10^0 )=

(

6. 234 × 10 −^24

)

(4. 44 × 10 −^3 )m^2

=2. 771 × 10 −^26 m^2

σ(θ=30^0 )=

(

8. 016 × 10 −^26

)

(4. 44 × 10 −^3 )m^2

=3. 563 × 10 −^28 m^2.

If we know the atomic densityDand thicknesstof the target material, we

can estimate the number of particles scattered at an angleθby

Nθ=N 0

 σ(θ)Dt,

whereN 0 is the number of incident particles. Therefore, for the two given

angles we get

N 100 =

(

106

)(

2. 771 × 10 −^26

)(

6 × 1028

)(

10 −^6

)

≈ 1663 s−^1

N 100 =

(

106

)(

3. 563 × 10 −^28

)(

6 × 1028

)(

10 −^6

)

≈ 21 s−^1.

Since the detector is only 60% efficient, therefore the count rate recorded at

the two angles will be

C 100 ≈ (1663)(0.6) = 998s−^1

C 300 ≈ (21)(0.6) = 13s−^1.