524 9 Particle Physics – I
9.76 p=^0.^3 BR (1)
wherepis in GeV/c,Bin Tesla andRin metres.
Nowθ≈L/R= 0. 3 BL/p (2)
whereLis the length of the magnet,θis the angle of deflection,Ris the radius
of curvature of the circular arc of the path in the magnetic field, andlis the
length of the straight path from the slits (Fig. 9.11).
Δθ=s/l= 0. 3 BLΔp/p^2
Δp/p= 1 / 100
S= 0. 3 × 1. 2 × 1. 5 × 10 ×(1/100)(1/25)
= 2. 16 × 10 −^3 m= 2 .16 mm
Fig. 9.11
9.3.11 Betatron..........................................
9.77ΔT=eΔφ/Δt= 1. 6 × 10 −^19 ×50 J
=50 eV
9.78 (a) IfNis the number of revolutions then
(2πRN)(4f)=c
The factor 4 arises due to the fact that the duty cycle is over a quarter of a
period.
N=c/ 8 πRf= 3 × 108 / 8 π× 0. 9 × 50 = 2. 65 × 105
(b) Radius,R= 0 .9m
Tmax=BRec= 1. 2 × 0. 9 × 1. 6 × 10 −^19 × 3 × 108
= 5. 184 × 10 −^11 J
= 5. 184 × 10 −^11 / 1. 6 × 10 −^13 =324 MeV
(c) The average energy gained per revolution
ΔT=Tmax/N=324 MeV/ 2. 65 × 105
= 122. 2 × 10 −^5 MeV
= 1 .222 keV
9.3.12 Cyclotron.........................................
9.79 The resonance condition is
ω=qB/m