1000 Solved Problems in Modern Physics

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 σ

9.57 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