530
9 Particle Physics – I
9.98 (a) At injection γ = 1 +
10
0.511
= 20.57
β = 0.9988
At extraction γ = 1 +
5,000
0.511
= 9, 786
β ≈ 1
Initial frequency f 1 =
βc
2πr
=
(0.9988) × 3 × 10
8
2π × 15
c = 3.1809 Mc
Final frequency f 2 =
1 × 3 × 10
8
2π × 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= E f − E i = 5,000 − 10 = 4,990 MeV
Energy gain per turn = 1 keV
∴ Number of turns, n =
4,990
10 −3 = 4.99 × 10
6
(c) The period of revolution
T 0 = 1/ f = 1/3.18 × 10
6
= 3.14 × 10
−7 s
Time between injection and extraction is
T = nT 0 = 4.99 × 10
6
× 3.14 × 10
−7
= 1.567 s
(d) The total distance traveled by the electron is
d = 2π rn = 2π × 15 × 4.99 × 10
6
= 4.7 × 10
8 m = 4.7 × 10
5 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 is ΔE = (50–2)/96 = 0.5 MeV/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 × 10
10
/2 × 2 × 10
8 ) 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
l f = βc/2 f = 0.313 × 3 × 10
10
/2 × 2 × 10
8
= 23.48 cm
(b) To produce 80 MeV protons, number of additional tubes required is (80 −
50)/0.5 = 60
9.100 (a) Beam current i = q/t = 50 × 5 × 10
11
× 1.6 × 10
−19
= 4 × 10
−6 amp
= 4μA
9 Particle Physics – I
9.98 (a) At injection γ = 1 +
10
0.511
= 20.57
β = 0.9988
At extraction γ = 1 +
5,000
0.511
= 9, 786
β ≈ 1
Initial frequency f 1 =
βc
2πr
=
(0.9988) × 3 × 10
8
2π × 15
c = 3.1809 Mc
Final frequency f 2 =
1 × 3 × 10
8
2π × 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= E f − E i = 5,000 − 10 = 4,990 MeV
Energy gain per turn = 1 keV
∴ Number of turns, n =
4,990
10 −3 = 4.99 × 10
6
(c) The period of revolution
T 0 = 1/ f = 1/3.18 × 10
6
= 3.14 × 10
−7 s
Time between injection and extraction is
T = nT 0 = 4.99 × 10
6
× 3.14 × 10
−7
= 1.567 s
(d) The total distance traveled by the electron is
d = 2π rn = 2π × 15 × 4.99 × 10
6
= 4.7 × 10
8 m = 4.7 × 10
5 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 is ΔE = (50–2)/96 = 0.5 MeV/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 × 10
10
/2 × 2 × 10
8 ) 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
l f = βc/2 f = 0.313 × 3 × 10
10
/2 × 2 × 10
8
= 23.48 cm
(b) To produce 80 MeV protons, number of additional tubes required is (80 −
50)/0.5 = 60
9.100 (a) Beam current i = q/t = 50 × 5 × 10
11
× 1.6 × 10
−19
= 4 × 10
−6 amp
= 4μA