# Chapter 6: Particle Accelerators¶

## Example 6.2.1, Page 264¶

In :
#Variable declaration
q = 1;  #   Number of proton,
V = 800;    #  Voltage applied to the dome, kV

#Calculations
E = q*V; # The kinetic energy of proton,keV

#Result
print "The kinetic energy of proton : %d keV"%E

The kinetic energy of proton : 800 keV


## Example 6.3.1, Page 265¶

In :
#Variable declaration
q = 1;  #   Number of proton,
V = 7;    #  Voltage applied to the dome, MV

#Calculations
E = q*V; # The kinetic energy of proton,MeV

#Result
print "The kinetic energy of proton : %d MeV"%E

The kinetic energy of proton : 7 MeV


## Example 6.3.2, Page 265¶

In :
#Variable declaration
V = 5; # Voltage of accelerator, MV
# Declare three cells [for three reactions]: Page no. : 133
R1 = [[0,0],[0,0],[0,0]]
R2 = [[0,0],[0,0],[0,0],[0,0],[0,0],[0,0],[0,0],[0,0],[0,0],[0,0]]
# Enter data for first cell [Reaction]
R1 = "p";
R1 = 1;
R1 = 'd';
R1 = 1;
R1 = "He";
R1 = 2;

#Calculations&Results
E_p =  R1*V
E_d =  R1*V
E_He =  R1*V
# Enter data for second cell [Reaction]
R2 = "p"
R2 = 1
R2 = "N"
R2 = 14
R2 = "O"
R2 = 15
R2 = "y"
R2 = 0
R2 = "d"
R2 = 1
R2 = "n"
R2 = 0
R2 = "He"
R2 = 3
R2 = "C"
R2 = 13
R2 = "He"
R2 = 4
R2 = "C"
R2 = 12
print "Protons energy   = -%d MeV \n Deuterons energy   = -%d MeV  \n Double charged He-3  = -%d MeV"%(E_p, E_d, E_He)
print " Possible reaction at these energies are"
print " %s + %s[%d] --->   %s[%d]+ %s"%(R2,R2,R2,R2,R2,R2)
print " %s + %s[%d] --->   %s[%d] + %s "%(R2,R2,R2,R2,R2,R2)
print " %s[%d] +%s[%d]  --->   %s[%d]+ %s"%(R2,R2,R2,R2,R2,R2,R2)
print " %s[%d] + %s[%d]  --->  %s[%d] +%s"%(R2,R2,R2,R2,R2,R2,R2)

Protons energy   = -5 MeV
Deuterons energy   = -5 MeV
Double charged He-3  = -10 MeV
Possible reaction at these energies are
p + N --->   O+ y
d + N --->   O + n
He +C  --->   O+ n
He + C  --->  O +n


## Example 6.4.1, Page 266¶

In :
#Variable declaration
q = 2;  #   Number of proton,
V = 15;    #  Voltage applied to the dome, MV

#Calculations
E = q*V; # The kinetic energy of proton,MeV

#Result
print "The kinetic energy of proton : %d MeV"%E

The kinetic energy of proton : 30 MeV


## Example 6.5.1, Page 265¶

In :
#Variable declaration
v = 2.999999997e+08;  #  Velocity of the electron, m/s
c = 3e+08;  # Velocity of light,m/s

#Calculations
D = c-v; # difference between electron's speed and speed of light,m/s

#Result
print "The difference between electron speed and speed of light : %3.1f m/s"%D

The difference between electron speed and speed of light : 0.3 m/s


## Example 6.5.2, Page 268¶

In :
import math

#Variable declaration
f = 200e+06;  #  Frequency of applied the voltage, Hz
V_0 = 750e+03;  # Applied potential difference, V
q = 1.6e-019; # Charge of proton, C
m = 1.67e-027; # Mass of proton, Kg
n_1 = 1; # For first tube

#Calculations
L_1 = math.sqrt(2*n_1*q*V_0/m)/(2*f);   # Length of the first tube, m
n_n = 128; #  For last tube
L_n = 1/(2*f)*math.sqrt(2*n_n*q*V_0/m); # Length of the last tube,m

#Result
print "Length of the first tube = %4.2f m \n Length of the last tube = %4.2f m "%(L_1,L_n)

Length of the first tube = 0.03 m
Length of the last tube = 0.34 m


## Example 6.5.3, Page 269¶

In :
#Variable declaration
K_E = 1.17; # Kinetic energy of the electron, MeV
E_r = 0.511; # Rest mass energy of the electron, MeV

#Calculations
v = (1-1/(K_E/E_r+1)**2); # Velocity of the electron, m/s

#Result
print "Velocity of the electron : %4.2fc"%v

Velocity of the electron : 0.91c


## Example 6.7.1, Page 270¶

In :
import math

#Variable declaration
V = 20e+03; # Potential difference across the dees, V
r = 0.28; # Radius of the dees, m
B = 1.1; # Magnetic field, tesla
q = 1.6e-019; # Charge of the proton, C
m = 1.67e-027; # Mass of the proton, Kg

#Calculations
E_max = B**2*q**2*r**2/(2*m*1.6e-013); # Maximnum energy acquired by protons,MeV
f = B*q/(2*math.pi*m*10**06); # Frequecy of the oscillator,MHzmath.
N = E_max*1.6e-013/(q*V); # Number of revolutions,

#Result
print "Maximum energy acquired by proton   =  %4.2f MeV \n Frequency of the oscillator = %4.2f MHz \n Number of revolutions =  %d revolutions "%(E_max,f,N)

Maximum energy acquired by proton   =  4.54 MeV
Frequency of the oscillator = 16.77 MHz
Number of revolutions =  227 revolutions


## Example 6.7.2, Page 271¶

In :
import math

#Variable declaration
B= 2.475; # Magnetic field, tesla
q = 1.6e-019; # Charge of the deutron, C
m = 2*1.67e-027; # Mass of the deutron, Kg

#Calculations
f = B*q/(2*math.pi*m*10**06); # Frequency of the deutron,MHz

#Result
print "Frequency of the deutron:  %4.2f MHz "%f

Frequency of the deutron:  18.87 MHz


## Example 6.7.3, Page 271¶

In :
import math

#Variable declaration
q = 1.6e-019; # Charge of the proton, C
r = 0.60; # radius of the dees, m
m = 1.67e-027; # Mass of the proton, Kg
f = 10**6; # Frequecy of the proton,Hz

#Calculations
B = 2*math.pi*m*f/q; # Magnetic field applied to cyclotron, tesla

#Result
print "Magnetic field applied to cyclotron :  %6.4f tesla "%B

Magnetic field applied to cyclotron :  0.0656 tesla


## Example 6.7.4, Page 272¶

In :
import math

#Variable declaration
q = 1.6e-019; # Charge of the proton, C
m = 1.67e-027; # Mass of the proton, Kg
B = 1.4; # Magnetic field , tesla

#Calculations
f = B*q/(2*math.pi*m*10**06); # Frequency of the applied field, tesla

#Result
print "Frequency of the applied field :  %4.2f MHz"%f

Frequency of the applied field :  21.35 MHz


## Example 6.8.1, Page 273¶

In :
import math

#Variable declaration
e = 1.6e-019 ; # Charge of an electron, C
f = 60; # Frequency of variation magnetic field, Hz
B_0 = 1; # Magnetic field , tesla
r_0 = 1; # Radius of doughnut, m

#Calculations
E = 4*e*2*math.pi*f*r_0**2/(1.6e-019); # Energy gained by electron per turn, eV
E_g = round(E)

#Result
print "Energy gained by electron per turn:  %d eV"%E_g

Energy gained by electron per turn:  1508 eV


## Example 6.9.1, Page 273¶

In :
#Variable declaration
K = 500; # Kinetic energy of the proton, MeV
E_r = 938.; # Rest mass energy of the proton, MeV

#Calculations
R_f = E_r/(K+E_r); # The ratio of highest to the lowest frequency,

#Result
print "The ratio of highest to the lowest frequency :  %4.2f "%R_f

The ratio of highest to the lowest frequency :  0.65


## Example 6.9.2, Page 274¶

In :
#Variable declaration
E_k = 1200; # Kinetic energy of the proton, MeV
q = 7; # Number of proton in nitrogen
E_r = 13040 #  Rest mass energy of the electron, MeV

#Calculations
E = (E_k+E_r)*1.6e-013; # Total energy,j
c = 3e+08; # Velocity of light, m/s
R_w_B = q*1.6e-019*c**2/E; # Ratio of w/B, m^2/W

#Result
print "The ratio of w/B :  %4.2e m^2/W "%R_w_B

The ratio of w/B :  4.42e+07 m^2/W


## Example 6.10.1, Page 274¶

In :
#Variable declaration
q = 1.602e-019; # Charge of an electron, C
r = 0.28; # Radius of stable orbit,m
E = 70*1.6e-013;  #  Energy of the electron, j
c = 3e+08; # Velocity of light, m/s

#Calculations
B = E/(q*r*c); # Magnetic field, T

#Result
print "The magnetic field of the electron :  %4.2f T"%B

The magnetic field of the electron :  0.83 T


## Example 6.10.2, Page 275¶

In :
#Variable declaration
c= 3e+08; # Speed of light in vacuum, m/s
q = 1.602e-019; # Charge on proton, coulomb
amu = 931;    # Energy equivalent of 1 amu, MeV
m = 938;  #  Rest mass of a proton, MeV
KE = 12e+03;    # Kinetic energy of proton, MeV
B = 1.9; # Magnetic field, T

#Calculations
E = m + KE;    # Total energy of proton, MeV
# As E = m*amu, solving for m, the mass of proton
m = E/amu*1.672e-027;    # Proton mass in motion, kg
v = 0.9973*c; # Velocity of the proton, m/s
r = m*v/(B*q); # Radius of the proton, m

#Result
print "Radius of the proton orbit :  %4.2f m"%r

Radius of the proton orbit :  22.84 m