1000 Solved Problems in Modern Physics

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 × 1026 Js−^1 =

3. 83 × 1026

1. 6 × 10 −^13

MeV s−^1

= 2. 39 × 1039 MeV s−^1

Number ofα′s produced,Nα=

2. 39 × 1039

26. 72

= 8. 94 × 1037 s−^1

As the number of neutrinos produced is double the number ofα’s,

Nν= 2 Nα= 1. 788 × 1038 s−^1

Number of neutrinos received per square metre on earth’s surface, that is flux

φ=

Nυ

4 πr^2

=

1. 788 × 1038

( 4 π)

(

1. 5 × 1011

) 2 =^6.^33 ×^1014 m−^2 s−^1

10.75 Using the formulaE=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 energiesE 1 andE 2 where

E 2 >E 1 >>m,

Δt=

d

2 c

(

mc^2

) 2

[

1

E^21

−

1

E 22

]

SubstitutingΔt=4s,E 1 =5 MeV andE 2 =15 MeV, dc = 17 × 104 ×

3. 15 × 107 seconds, and solving formc^2 we findmc^2 = 6. 48 × 10 −^6 MeV=

6 .5eV

10.76 (a)Q/e=+^1 ∵ΔQ=^0
(b)B= 0 ∵ΔB= 0
(c)lμ= 0 ∵Δlμ= 0
(d)T=^1 / 2 ∵ΔT 3 =± 1 / 2
(e)s=± 1 ∵ΔS=± 1
(f) spin=0or1∵spin ofπ^0 ,μandνare 0,^1 / 2 and^1 / 2 respectively
(g) Boson ∵on right hand side there are two fermions (μ, ν) and
one boson (π)