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.