1000 Solved Problems in Modern Physics

10.3 Solutions 563

(b) Each photon will carry 67.5 MeV

λ=

1241

67. 5 × 106

nm= 18. 385 × 10 −^6 nm= 1. 84 × 10 −^14 m

10.3.2 StrongInteractions.................................

10.17 (a)τ=



Γ

=

c

Γc

=

197. 3 × 10 −^15 (MeV−m)

(120MeV)(3× 108 m/s)

= 5. 5 × 10 −^24 s

γ= 1 +

200

1. 236

∼= 163

β∼= 1

d=βcγτ= 1 × 3 × 108 × 163 × 5. 5 × 10 −^24 m

∼= 2. 7 × 10 −^13 m

∼= 0 .0003 nm

a distance which is much less than the resolution obtainable by the avail-

able techniques. The best resolution obtained in photographic emulsions

is only 1μm.

(b) Uncertainty Principle:

Γ.τ≥

τ≥



Γ

=

c

Γc

=

197 .3MeV−fm

6 .5(MeV)× 3 × 108 (m/s)

=

197. 3 × 10 −^15

19. 5 × 108

τ≥ 10

− 22

s

10.18π

++p→π++p (1)

π−+p→π−+p (2)

π−+p→π^0 +n (3)

The cross-section is proportional to the square of the matrix elementMif

connecting the initial and final states

Mif=

〈

ψf|H|ψi

〉

whereHis the isospin operator, and

σ∝

∣

∣Mif

∣

∣^2

As pion hasT=1 and protonT = 1 /2, the reactions can proceed either

throughI= 1 /2orI= 3 /2 channels. Designating the corresponding oper-

ators byH 1 andH 3 and the matrix elements for the reactions byM 1 andM 3 ,

we can write

M 1 =

〈

ψf

(

1

2

)∣∣

∣

∣H^1

∣

∣

∣

∣ψi

(

1

2

)〉

M 3 =

〈

ψf

(

3

2

)∣

∣

∣

∣H^3

∣

∣

∣

∣ψi

(

3

2

)〉

AsI-spin is conserved there is no operator connecting different isospin

states. Reaction (1) involves a pure state ofI= 3 / 2 ,I 3 =+ 3 /2.