1000 Solved Problems in Modern Physics

530 9 Particle Physics – I

9.98 (a) At injectionγ= 1 + 0.^10511 = 20. 57

β= 0. 9988

At extractionγ= 1 +^50 ,.^000511 = 9 , 786

β≈ 1

Initial frequencyf 1 =

βc

2 πr

=

(0.9988)× 3 × 108

2 π× 15

c= 3 .1809 Mc

Final frequencyf 2 =

1 × 3 × 108

2 π× 15

= 3 .1847 Mc

As the initial and final frequencies are nearly the same there is hardly

any need to change the R.F. frequency.

(b) Total energy gain=Ef−Ei= 5 , 000 − 10 = 4 ,990 MeV

Energy gain per turn=1keV

∴Number of turns,n=^410 ,^990 − 3 = 4. 99 × 106

(c) The period of revolution

T 0 = 1 /f= 1 / 3. 18 × 106 = 3. 14 × 10 −^7 s

Time between injection and extraction is

T=nT 0 = 4. 99 × 106 × 3. 14 × 10 −^7 = 1 .567 s

(d) The total distance traveled by the electron is

d= 2 πrn= 2 π× 15 × 4. 99 × 106

= 4. 7 × 108 m= 4. 7 × 105 km.

9.3.14 Linear Accelerator ................................

9.99 (a) As there are 97 drift tubes, there will be 96 gaps. The energy gain per

gap isΔE=(50–2)/ 96 = 0 .5MeV/gap

After crossing the first gap, the protons are still non-relativistic and

their velocity will be in the second drift tube will be

v 2 =(2T/m)^1 /^2 =c[2×(2+ 0 .5)/938]^1 /^2 = 0. 073 c

The length of the second tube

l 2 =v 2 / 2 f=(0. 073 × 3 × 1010 / 2 × 2 × 108 )cm

= 5 .48 cm

In the last tube v is calculated relativistically.

γ= 1 +T/m= 1 + 50 / 938 = 1. 053

β=(γ^2 −1)^1 /^2 /γ= 0. 313

lf=βc/ 2 f= 0. 313 × 3 × 1010 / 2 × 2 × 108 = 23 .48 cm

(b) To produce 80 MeV protons, number of additional tubes required is (80−

50)/ 0. 5 = 60

9.100 (a) Beam currenti=q/t= 50 × 5 × 1011 × 1. 6 × 10 −^19

= 4 × 10 −^6 amp

= 4 μA