518 9 Particle Physics – I
Atx ̄,p(x ̄)=(
σ√
2 π)− 1
(2)
Let the HWHM points be located at
x ̄−x=kσ (3)
Thenp(kσ)=(
σ√
2 π)− 1
exp(−k^2 /2) (4)
By definition,
p(kσ)
p(x ̄)=
1
2
=exp(
−
k^2
2)
(5)
where we have used (2) and (4). Taking logeon both sides of (5)
k^2 / 2 =ln 2= 0. 69315
∴k= 1. 1774
Therefore HWHM= 1. 177 σ9.57 Channel 298 corresponds to 661 keV energy (Fig. 9.7)
FWHM (full width at half maximum)= 316 − 281 =35 channels
HWHM (half width at half maximum)= 35 / 2 = 17 .5 channels
HWHM corresponds to (661 keV)(17.5)/ 298 = 38 .8keV
Now HWHM= 1. 177 σ
Thereforeσ=HWHM/ 1. 177
= 38. 8 / 1. 177 =33 keV
The coefficient of the energy determination=(33)(100)/ 661 =5%
Fig. 9.7Photo peak 661 keV
γ-rays
9.3.7 Cerenkov Counter
9.58 When a charged particle moves through a medium with velocity greater than
the light velocity in the medium then light is emitted known as Cerenkov radi-
ation, after the discoverer. The radiation is emitted around the surface of a
cone with its axis along the particle’s path and with a half-angle given by