Radiation Detector 2015 (SPA6309), Homework 2
© 2015 Teppei Katori
Name:
ID:
Problem 1 (5 points)
Supernova type IIa (core collapose supernova) produces significant amount of detectable
neutrinos and many people are waiting the next one within the Milky way. Here, the rate of type
IIa supernova is predicted 3 per century.
[1] Assuming poisson distribution, what is the probability to measure 0, 1, or 3 type IIa
supernovae in 100 years? (3 points)
[2] Next, from the observation, poeple cannot find any type IIa supernova in 100 years. What is
the upper limit of the rate of type IIa supernova observation with 90% confidential level? (2
points)
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In [1]: import numpy as np
import matplotlib.pylab as plt
a_list=[0,1,2,3,4,5,6,7,8]
b_list=[]
mu=3
for n in a_list:
b_list.append(3**n*exp(-mu)/math.factorial(n))
print "Prob. to detect",n," is ",b_list[n]
prob_bin = np.linspace(0,8,9)
plt.plot(prob_bin,b_list,marker='o',linestyle='-',color='grey',label='SN IIa'
plt.show()
Prob.
Prob.
Prob.
Prob.
Prob.
Prob.
Prob.
Prob.
Prob.
to
to
to
to
to
to
to
to
to
detect
detect
detect
detect
detect
detect
detect
detect
detect
0
1
2
3
4
5
6
7
8
is
is
is
is
is
is
is
is
is
0.0497870683679
0.149361205104
0.224041807655
0.224041807655
0.168031355742
0.100818813445
0.0504094067225
0.0216040314525
0.00810151179468
In [2]: Lo=-log(0.1)
print "rate of 0 type IIa supernova is ",Lo,"per 100 year"
rate of 0 type IIa supernova is
2.30258509299 per 100 year
Even after 100 years measurement, upper limit is 2.3 supernova per century (90%CL), so you
cannot eliminate the possibility to see supernova in next 100 years.
problem 2 (3 points)
During an experiment, array of 50 independent counters are monitoring background particles
entering the detector to reject them (=veto counters). Each of them have 99% efficiency to
detect particles, or 1% chance to fail to detect particles entering to the detector. An experiment
discard the data if 2 or more counters fail to reject background particles. How many percentage
of the data from this experiment will be useful?
Using binomial distribution, total probability is sum of 0 counter failure and 1 counter failure.
P = B(50; 50, 0.99) + B(49; 50, 0.99) = (0.99)50 +50 C49 (0.99)49 ⋅ 0.01 = 0.605 + 0.305 =
so 91% of data will be useful.
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problem 3 (2 points)
There is a series of measurements about x , where mean is μ and variance is σ 2 . The average of
n measurement is xAV and the expectation value of xAV is μ. These quantities have following
relationships.
xAV = Σni=1 xi ,
$\mu=$, and
σ 2 =< (x − μ)2 >
The variance of this measurement σs2 is defined by
σs2 =
1
Σn ( x
n−1 i=1 i
− xAV )2
Prove expectation value of σs2 is indeed σ 2 .
< σs2 >=
1
Σn
n−1 i=1
2
⟨[(xi − μ) − (xAV − μ)] ⟩
=
1
Σn
n−1 i=1
1 n
2
⟨[(xi − μ) − n Σj=1 (xj − μ)] ⟩
=
1
Σn
n−1 i=1
2
⟨(xi − μ) ⟩ −
=
nσ 2
n−1
−
2
Σ n Σn
n(n−1) i=1 j=1
⟨(xi − μ)(xj − μ)⟩ +
=
nσ 2
n−1
−
1
Σ n Σn
n(n−1) i=1 j=1
⟨(xi − μ)(xj − μ)⟩
=
nσ 2
n−1
−
1
Σ n Σn
n(n−1) i=1 j=i
=
2
nσ
n−1
−
1
Σn
n(n−1) i=1
=
nσ 2
n−1
−
nσ 2
n(n−1)
2
Σn Σ n
n(n−1) i=1 j=1
⟨(xi − μ)(xj − μ)⟩ −
2
⟨(xi − μ) ⟩
= σ2
Useful constants and equations
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⟨(xi − μ)(xj − μ)⟩ +
1
Σn ( x
n(n−1) ⟨[ j=1 j
1
Σn Σn
n(n−1) i=1 j≠i
1
Σn
n2 (n−1) i=1
− μ)] ⟩
n
⟨[Σj=1 (xj − μ)] ⟩
2
⟨(xi − μ)(xj − μ)⟩
2
Δν(133 Cs)hf s = 9192631770 Hz
h = 6.626 × 10−34 J ⋅ s
c = 2.99792458 × 108 m/s
e = 1.602 × 10−19 C
NA = 6.022 × 1023
kB = 1.380 × 10−23 J/K
ℏ ⋅ c = 197 MeV ⋅ f m
mp = 1.007u
mn = 1.009u
1u = 931.5 MeV
GN = 6.674 × 10−11 N ⋅ m2 /kg2
M⊙ = 1.989 × 1030 kg
R⊙ = 6.961 × 108 m
L⊙ = 3.846 × 1026 J/s
1pc = 3.26lyr
Semi-empirical mass formula of the binding energy
B(Z, A) = av A − as A2/3 − ac
Z(Z−1)
A1/3
− asym
(A−2Z)2
A
±
ap
A3/4
Orbital angular momentum, L=0, 1, 2, 3... are L=s, p, d, f,...
The ordering of nuclear shell
1s1/2 : 1p3/2 : 1p1/2 : 1d5/2 : 2s1/2 : 1d3/2 : 1f7/2 : 2p3/2 : 1f5/2 : 2p1/2 : 1g9/2 : 2d5/2 : 1g
Periodic table
H, He, Li, Be, B, C, N, O, F, Ne, Na, Mg, Al, Si, P, S, Cl, Ar, K, Ca, Sc, Ti, V, Cr, Mn, Fe,
Co, Ni, Cu, Zn, Ga, Ge, As, Se, Br, Kr, Rb...
relativistic formula
2 c4 + p2 c‾
2 = mc2 γ , Δt ′ = γΔt , ΔL ′ = γ −1 ΔL , γ = (1 − β 2 )−1/2 , β = v/c
E = √‾m‾‾‾‾‾‾‾‾‾‾
Nuclear mass
M (AZ X) = Z mp + N mn − B
Rutherford scattering formula
dσ
dΩ
zZe
1
= ( 4πϵ
0 ) ( 4Tα )
2
2
2
1
sin4
1
= ( zZℏc
137 ) ( 4Tα )
2
2
θ
2
1
sin4
Radius of nucleus
R ∼ Ro A1/3 , where Ro 1.20 f m
Fermi's 4-factor formula (FFFF , actually 6 factors)
θ
2
k = ηϵpf (1 − lf )(1 − lt )
Binomial distribution
B(x, n, p, q) = n Cx px q(n−x)
Poisson distribution
P(x, μ) =
μx −μ
e
x!
Gaussian distribution G(x, μ, σ 2 )
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=
1
√2πσ 2
exp [
(x−μ)2
2σ 2
]
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