November 2023 Nuclear Human Resources Initiative Practicum
Kentaro Hirose

Derivation of evaporation residual
nucleus cross section

Nuclei produced in the evaporation process following nuclearfusion
ELAB (MeV)

3416S

103

Cross-sectionalarea(mb)(mb)
Cross section

169
69Tm

Compound nucleus
203
85At

Nuclear fission

n202
85At

Nuclear fission

n201
85At

Nuclear fission

n135 140 145 150 155 160 165 170 175 180

102

Nuclear fissionnDisintegrates by fission
FissionResidual nucleus
Evaporation
Residue101
200

10At(3n)0199

At(4n)

199

Po(p3n)10-1

10-220085AtFusion
Fusion198

At(5n)

198

Po(p4n)200

Po(p2n)

201

At(2n)

10-3
110 115 120 125 130 135 140 145 150

19985At

Residual nucleus

About a few percent of thecomposite nuclei survive as
evaporated residual nuclei

Ec.m. (MeV)

34SBeam＝10 pnA

Cross-sectional area = 1.0 mb
Target thickness ＝100μg/㎝2
RMS Transport Efficiency ＝10％
→ 800 events in 1 hourCross-sectional area・・・Ease of reaction
3416S

169
69Tm

Neutronemission

199
85At

The nuclear beam 34S is applied to the target
nucleus 169Tm, The number of residual nuclei
199Atproduced is,

Number of 34S beam nuclei to be hit

Number of target nuclei 169Tm to be hit
is proportional.

Density of target nuclei[1/cm3]
Number of 199
85At
[1]

TargetNumber ofthickness
incident[cm]nuclei [1]𝑁𝑁 19985At ∝ 𝑁𝑁𝐵𝐵 ⋅ 𝑛𝑛 ⋅ 𝑡𝑡 = 𝑁𝑁𝐵𝐵 ⋅ 𝑁𝑁𝑇𝑇

𝑁𝑁 19985At = 𝜎𝜎 ⋅ 𝑁𝑁𝐵𝐵 ⋅ 𝑁𝑁𝑇𝑇

Areal density of target[1/cm2]=“Target thickness”

Let 𝜎𝜎 be the proportionality constant.

𝑁𝑁 19985At [1]2 σ is the dimension of area.
=𝜎𝜎[cm] A value specific to the reaction.
𝑁𝑁𝐵𝐵 [1] ⋅ 𝑁𝑁𝑇𝑇 [1⁄cm2 ](*) 1[cm2] is too large a unit
Use 10-28[m2] = 1[b] (b:barn).

Carbon foilCounting 𝑁𝑁 199
85At

169Tm

(1/2)

𝑁𝑁 19985At
𝜎𝜎 =𝑁𝑁𝐵𝐵 ⋅ 𝑁𝑁𝑇𝑇

199At

34S

199

At 𝑞𝑞+ (Orbit of charge state q)

Yield

If q=16 is set to pass through the RMS, q=15,17, etc. will not reach the terminal detector.

The 199At ratio for q=16 is 18%

18% of the 199At generated reaches the end detector

Charge

199
𝑁𝑁85At199
199
Count the number 𝑁𝑁 85At of At (2/2)𝜎𝜎 =𝑁𝑁𝐵𝐵 ⋅ 𝑁𝑁𝑇𝑇19885At
19985At

20085At𝐸𝐸𝛼𝛼 = 6.753MeV(97%)
𝑇𝑇1⁄2 = 4.2s

19483Bi𝐸𝐸𝛼𝛼 = 6.465MeV(57%)
𝑇𝑇1⁄2 = 43s

196
83Bi

𝐸𝐸𝛼𝛼 = 6.643MeV(90.0%)𝑇𝑇1⁄2 = 7.03s

19985At
20085At

19885At

195
83Bi

charge state
161718198
At 12.38 11.65 11.00
199
At 12.44 11.71 11.06
200
At 12.50 11.76 11.11

No difference in m/q = orbits are close
Not only 199At reaches the detector

199At has a 90% probability of emitting an alpha ray of 6.643 MeV.

Count the number from the spectrum measured by the detector.

Eα𝑁𝑁 6.643MeV = 𝑁𝑁 199
85At × 0.18 × 0.90

Probability of reaching the detector

Probability of emitting an alpha ray of 6.643 MeV

𝑁𝑁 199
85At
Count (number of incident beams 𝑁𝑁𝐵𝐵 )×(number of targets 𝑁𝑁𝑇𝑇 ) 𝜎𝜎 =𝑁𝑁𝐵𝐵 ⋅ 𝑁𝑁𝑇𝑇169Tm

1.E+06[mb/msr]

34S

1.E+04
1.E+02

𝑑𝑑𝜎𝜎𝑅𝑅𝑅𝑅𝑅𝑅𝑅
𝑑𝑑Ω

2.86 mbarn/msr(@ θ=45°)

1.E+00Collimator
PIN diode（D iameter 8mm）
1.E-02
Distance 120mm、Diameter 8mm 0 ΔΩ=3.49 msr
𝑑𝑑𝜎𝜎𝑅𝑅𝑅𝑅𝑅𝑅𝑅⋅ ΔΩ ⋅ 𝑁𝑁𝐵𝐵 ⋅ 𝑁𝑁𝑇𝑇𝑑𝑑Ω
1𝑑𝑑𝜎𝜎𝑅𝑅𝑅𝑅𝑅𝑅𝑅 ΔΩ
9.98==[mbarn]𝑑𝑑Ω𝑁𝑁𝐵𝐵 ⋅ 𝑁𝑁𝑇𝑇
𝑁𝑁𝑒𝑒𝑒𝑒𝑒𝑒 𝑁𝑁𝑒𝑒𝑒𝑒𝑒𝑒

𝑁𝑁𝑒𝑒𝑒𝑒𝑒𝑒 =306090120150180

Scattering angle(deg)Variables NB and NT can be erased.
You don't need to know directly how much of the beam
you hit!
No need to know the variation of NT due to irradiation!
Furthermore, these errors are not reflected in the results!

𝑁𝑁 199𝑑𝑑𝜎𝜎𝑅𝑅𝑅𝑅𝑅𝑅𝑅 ΔΩ � 𝑁𝑁 199
9.98
𝑁𝑁 6.643MeV
85At
85At
𝜎𝜎 ===
𝑁𝑁𝐵𝐵 ⋅ 𝑁𝑁𝑇𝑇
𝑁𝑁𝑒𝑒𝑒𝑒𝑒𝑒𝑑𝑑Ω
0.18 × 0.90
𝑁𝑁𝑒𝑒𝑒𝑒𝑒𝑒mbarn(*) Peak at 6.643 MeV measured with Si detector , cross sections can bederived by simply counting the number of elastic scattering peaks
measured with a PIN diode.