582
10 Particle Physics – II
|N , π, 1/2, −1/2 = −
2
3
pπ
−
+
1
3
nπ
0
Therefore,
ω
Λ → pπ
−
ω
Λ → nπ 0
=
2
3
2
1
3
2
= 2
10.74
L = 3.83 × 10
26 Js
−1
=
3.83 × 10
26
1.6 × 10 −13 MeV s
−1
= 2.39 × 10
39 MeV s
−1
Number of α
s produced, N α =
2.39 × 10
39
26.72
= 8.94 × 10
37 s
−1
As the number of neutrinos produced is double the number of α’s,
N ν = 2N α = 1.788 × 10
38 s
−1
Number of neutrinos received per square metre on earth’s surface, that is flux
φ =
N υ
4πr 2 =
1.788 × 10
38
(4π)
1.5 × 10 11
2 = 6.33 × 10
14 m
−2 s
−1
10.75 Using the formula E = mc
2
γ , the speed of antineutrinos of mass m
v = c
1 −
m
2 c
4
E 2
1 / 2
The time taken to travel to earth
t =
d
v
=
d
c
1 −
m
2 c
4
E 2
− 1 / 2 ∼ =
d
c
1 +
1
2
m
2 c
4
E 2
The difference in travel time for two neutrinos of energies E 1 and E 2 where
E 2 > E 1 >> m,
Δt =
d
2c
mc
2
2
1
E
2
1
−
1
E
2
2
Substituting Δt = 4 s, E 1 = 5 MeV and E 2 = 15 MeV,
d
c
= 17 × 10
4
×
3.15 × 10
7 seconds, and solving for mc
2 we find mc
2
= 6.48 × 10
−6 MeV =
6.5 eV
10.76 (a) Q/e = +1 ∵ ΔQ = 0
(b) B = 0
∵ ΔB = 0
(c) l μ = 0
∵ Δl μ = 0
(d) T = 1 / 2
∵ ΔT 3 = ±1/2
(e) s = ±1
∵ ΔS = ±1
(f) spin = 0 or 1 ∵ spin of π
0
, μ and ν are 0, 1 / 2 and 1 / 2 respectively
(g) Boson
∵ on right hand side there are two fermions (μ, ν) and
one boson (π)
10 Particle Physics – II
|N , π, 1/2, −1/2 = −
2
3
pπ
−
+
1
3
nπ
0
Therefore,
ω
Λ → pπ
−
ω
Λ → nπ 0
=
2
3
2
1
3
2
= 2
10.74
L = 3.83 × 10
26 Js
−1
=
3.83 × 10
26
1.6 × 10 −13 MeV s
−1
= 2.39 × 10
39 MeV s
−1
Number of α
s produced, N α =
2.39 × 10
39
26.72
= 8.94 × 10
37 s
−1
As the number of neutrinos produced is double the number of α’s,
N ν = 2N α = 1.788 × 10
38 s
−1
Number of neutrinos received per square metre on earth’s surface, that is flux
φ =
N υ
4πr 2 =
1.788 × 10
38
(4π)
1.5 × 10 11
2 = 6.33 × 10
14 m
−2 s
−1
10.75 Using the formula E = mc
2
γ , the speed of antineutrinos of mass m
v = c
1 −
m
2 c
4
E 2
1 / 2
The time taken to travel to earth
t =
d
v
=
d
c
1 −
m
2 c
4
E 2
− 1 / 2 ∼ =
d
c
1 +
1
2
m
2 c
4
E 2
The difference in travel time for two neutrinos of energies E 1 and E 2 where
E 2 > E 1 >> m,
Δt =
d
2c
mc
2
2
1
E
2
1
−
1
E
2
2
Substituting Δt = 4 s, E 1 = 5 MeV and E 2 = 15 MeV,
d
c
= 17 × 10
4
×
3.15 × 10
7 seconds, and solving for mc
2 we find mc
2
= 6.48 × 10
−6 MeV =
6.5 eV
10.76 (a) Q/e = +1 ∵ ΔQ = 0
(b) B = 0
∵ ΔB = 0
(c) l μ = 0
∵ Δl μ = 0
(d) T = 1 / 2
∵ ΔT 3 = ±1/2
(e) s = ±1
∵ ΔS = ±1
(f) spin = 0 or 1 ∵ spin of π
0
, μ and ν are 0, 1 / 2 and 1 / 2 respectively
(g) Boson
∵ on right hand side there are two fermions (μ, ν) and
one boson (π)