346 6 Special Theory of Relativity
6.56 TK=[(MD−MK)^2 −M^2 π]/ 2 MD=[(1865−494)^2 − 1402 ]/ 2 × 1865
= 498 .67 MeV
PK=(TK^2 + 2 MKTK)^1 /^2 =861 MeV/c
EK^2 =pK^2 +mK^2 =(0.861)^2 +(0.494)^2 = 0 .9853 GeV^2
EK= 0 .9926 GeV/c
Eπ^2 =pπ^2 +mπ^2 =(0.861)^2 +(0.140)^2 = 0 .7609 GeV^2
Eπ= 0 .8736 GeV
Mx=EK+Eπ= 1 .8662 GeV/c^2
It is aD^0 meson.
6.57 The mass of neutrino is zero. Applying conservation laws of energy and
momentum
Eμ+Eν=mπc^2 (1)
pμ=pν (2)
Multiplying (2) bycand squaring
c^2 pμ^2 =c^2 pν^2
OrEμ^2 −mμ^2 c^4 =Eν^2
OrEμ^2 −Eν^2 =mμ^2 c^4 (3)
Solve (1) and (3)
γμ=(m^2 π+m^2 μ)/ 2 mπmμ
βμ=(1− 1 /γμ^2 )^1 /^2 =(mπ^2 −mμ^2 )/(mμ^2 +mμ^2 )6.58EB+EC=mAc
(^2) (energy conservation) (1)
PB=PC (momentum conservation) (2)
Orc^2 PB^2 =c^2 PC^2 (3)
Using the relativistic equationsE^2 =c^2 p^2 +m^2 c^4 , (3) becomes
EB^2 −mB^2 c^4 =EC^2 −mC^2 c^4 (4)
EliminatingECbetween (1) and (4), and simplifying
EB=(mA^2 +mB^2 −mC^2 )c^2 / 2 mA (5)
6.59 K+→e++π◦+νe
The maximum energy of positron will correspond to a situation in which the
neutrino is at rest. In that case the total energy carried by electron will be
Ee(max)=(m^2 K+m^2 e−m^2 π 0 )/ 2 mK=(494^2 + 0. 52 + 1352 )/ 2 × 494 = 228. 5 MeV
∴Te(max)=228 MeV
6.60 Let the incident particle carry momentump 0. As the scattering is symmetrical,
each particle carries kinetic energyT/2 and momentumPafter scattering, and
makes an angleθ/2 with the incident direction.
Momentum conservation along the incident direction gives
p 0 =pcosθ/ 2 +pcosθ/ 2 = 2 pcosθ/2(1)
Or (T^2 + 2 Tmc^2 )^1 /^2 =2(T^2 / 4 + 2 T/ 2 mc^2 )^1 /^2 cosθ/2(2)
Squaring (2), and using the identity, cos^2 θ/ 2 =(1+cosθ)/ 2
We get the result cosθ=T/(T+ 4 mc^2 )