1000 Solved Problems in Modern Physics

512 9 Particle Physics – I

The integration of the expression

∫ mμ 2 c^2

0

E^2 e

(

1 − 4 Ee/ 3 mμc^2

)

dEe

gives (mμc^2 )^3 /48 so that the full expression for full width is

Γ=GF^2 (mc^2 )^5 /(c)^6192 π^3

which is identical with the one stated in Problem 9.5. The mean life time is

obtained fromτ=/Γas in the solution of 9.5.

9.34 (a) The charged pionsπ±decay by weak interactionsπ→μ+ν, and so

their mean lifetime is relatively longer, while the neutral pionπ^0 decays

viaπ^0 → 2 γ, electromagnetically and therefore their mean life time is

shorter.

(b) Theπ+andπ−are particle and antiparticle pair and they are expected to

have the same mass by the CPT theorem. On the other handΣ+andΣ−

hyperons are not particle-antiparticle. ActuallyΣ−−is the antiparticle of

Σ+.Σ+andΣ−are the members of the isospin triplet (Σ+,Σ^0 ,Σ−) and

because of difference in their charges can slightly differ in their mass sim-

ilar to the masses of neutron and proton for the isospin doublet of nucleon.

(c)Λ-hyperon decays viaΛ→p+π−orn+π^0 , the interaction is weak.

Similarly the hyperonΞ^0 decays viaΞ^0 →Λ+π^0 , which is also a weak

decay. In both the cases the lifetimes are relatively long on the nuclear

scale. On the other hand the decay ofΣ^0 -hyperon viaΣ^0 →Λ+γ is

electromagnetic. The explanation is the same as in (a)

9.35 Apply the formula

M^2 =mp^2 +mπ^2 +2(EpEπ−PpPπcosθ)(1)

Use the values:

mp= 0 .939 GeV,mπ= 0 .139 GeV,Pp= 0 .44 GeV

Pπ= 0 .126 GeV, Ep= 1 .036 GeV,Eπ= 0 .188 GeV,θ= 640 and find

M= 1 .114 GeV/c^2. The particle isΛhyperon

9.3.5 Ionization Chamber, GM Counter and Proportional Counters

9.36 The problem is based on the double source method for the determination of the
dead time of a G.M. counter. Two radioactive sources of comparable strength
are chosen. In all four counts are taken. First, the background rateBper second
is found out when neither source is present. One of the sources is placed in
a suitable position so that a high counting rate N 1 is registered. While this
counter is in the same position, the second source is placed by its side to get
the counting rate N 12. Finally, the first source is removed so that the second
counter above gives a count of N 2 .Ifn 1 ,n 2 and n 12 are the true counting rates
then we expect
(n 1 +B)+(n 2 +B)=n 12 +B