# Chapter11-Particle Accelerators¶

## Ex1-pg535¶

In [12]:
## Exa11.1 : : Page-535(2011)
#calculate The optimum number of stages in the accelerator and the ripple voltage
import math
V_0 = 10**5;        ## Accelerating voltage, volts
C = 0.02e-006;        ## Capacitance, farad
I = 4*1e-003;            ## Current, ampere
f = 200.;            ## Frequency, cycles per sec
n = math.sqrt (V_0*f*C/I);    ## Number of particles
delta_V = I*n*(n+1.)/(4.*f*C);
print'%s %.2f %s'%("\nThe optimum number of stages in the accelerator = ", n,"");
print'%s %.2f %s'%("\nThe ripple voltage = ", delta_V/1e+003,"kV");

## Result
## The optimum number of stages in the accelerator = 10
## The ripple voltage = 27.5 kV

The optimum number of stages in the accelerator =  10.00

The ripple voltage =  27.50 kV


## Ex2-pg536¶

In [11]:
## Exa11.2 : : Page-536 (2011)
#calculate The charging current and The rate of rise of electrode potential
import math
s = 15.;        ## Speed, metre per sec
w = 0.3;        ## Width of the electrode, metre
E = 3e+06;        ## Breakdown strength, volts per metre
eps = 8.85e-12;    ## Absolute permitivity of free space, farad per metre
C = 111e-12;        ## Capacitance, farad
i = round (2*eps*E*s*w*10**6);    ## Current, micro ampere
V = i/C*10**-12;            ## Rate of rise of electrode potential, mega volts per sec
print'%s %.2f %s %.2f %s '%("\nThe charging current =",i," micro-ampere"and " \nThe rate of rise of electrode potential = ",V," MV/sec");

## Result
## The charging current = 239 micro-ampere
## The rate of rise of electrode potential = 2.15 MV/sec

The charging current = 239.00
The rate of rise of electrode potential =  2.15  MV/sec


## Ex3-pg536¶

In [10]:
## Exa11.3 : : Page-536 (2011)
#calculate The length of the final drift tube and The kinetic energy of the injected protons
import math
import scipy
from scipy import integrate

f = 200.*10**6;        ## Frequency of the accelerator, cycle per sec
M = 1.6724e-27;        ## Mass of the proton, Kg
E = 45.3*1.6e-13;        ## Accelerating energy, joule
L_f = round (1./f*math.sqrt(2.*E/M)*100.);    ## Length of the final drift tube, centi metre
L_1 = 5.35*10**-2;                ## Length of the first drift tube, metre
K_E = (1./2.*M*L_1**2.*f**2.)/1.6e-13;    ## Kinetic energy of the injected proton, MeV
E_inc = E/1.6e-13-K_E;        ## Increase in energy, MeV
q = 1.6e-19;                ## Charge of the proton, C
V = 1.49e+06;            ## Accelerating voltage, volts
N = E_inc*1.6e-13/(q*V);   ## Number of drift protons
def fun(n):
y=n**0.5
return y

l2=scipy.integrate.quad(fun,0,N)
L = 1./f*math.sqrt(2.*q*V/M)*l2[0];    ## Total length of the accelerator, metre
print'%s %.2f %s %.2f %s %.2f %s '%("\nThe length of the final drift tube = ",L_f," cm"and"\nThe kinetic energy of the injected protons = ",K_E," MeV"and"\nThe total length of the accelerator = ",L," metre");

## Result
## The length of the final drift tube = 47 cm
## The kinetic energy of the injected protons = 0.60 MeV
## The total length of the accelerator = 9.2 metre

The length of the final drift tube =  47.00
The kinetic energy of the injected protons =  0.60
The total length of the accelerator =  9.25  metre


## Ex5-pg536¶

In [9]:
## Exa11.5 : : Page-536 (2011)
#find energy of the emergine deutron and frequency of the dee voltage
import math
B = 1.4;        ## Magnetic field, tesla
R = 88e-002;        ## Radius of the orbit, metre
q = 1.6023e-019;                ## Charge of the deutron, C
M_d = 2.014102*1.66e-27;        ## Mass of the deutron, Kg
M_He = 4.002603*1.66e-27;        ## Mass of the He ion, Kg
E = B**2*R**2*q**2/(2*M_d*1.6e-13);        ## Energy og the emerging deutron, mega electron volts
f = B*q/(2.*math.pi*M_d)*10**-6;        ## Frequency of the deutron voltage, mega cycles per sec
B_He = 2*math.pi*M_He*f*10**6/(2*q);    ## Magnetic field required for He(++) ions, weber per square metre
B_change = B-B_He;        ## Change in magnetic field, tesla
print'%s %.2f %s %.2f %s %.2f %s '%("\nThe energy of the emerging deutron = ",E," MeV" and "\nThe frequency of the dee voltage = ",f,"MHz" and"\nThe change in magnetic field = ",B_change," tesla");

## Result
## The energy of the emerging deutron = 36.4 MeV
## The frequency of the dee voltage = 10.68 MHz
## The change in magnetic field = 0.01 tesla

The energy of the emerging deutron =  36.42
The frequency of the dee voltage =  10.68
The change in magnetic field =  0.01  tesla


## Ex6-pg537¶

In [8]:
## Exa11.6: : Page-537 (2011)
#calculate effective reduction in magnetic field and charge in orbit radius
import math
K_E = 7.5*1.6023e-13;        ## Kinetic energy, joule
r = 0.51;                    ## Radius of the proton's orbit, metre
E = 5*10**6;                ## Electric field, volts per metre
m = 1.67e-27;            ## Mass of the proton, Kg
q = 1.6023e-19;                ## Charge of the proton, C
v = math.sqrt(2.*K_E/m);        ## Velocity of the proton, metre per sec
B_red = E/v;                ## The effective reduction in magnetic field, tesla
B = m*v/(q*r);            ## Total magnetic field produced, tesla
r_change = r*B_red/B;        ## The change in orbit radius, metre
print'%s %.2f %s %.2f %s '%("\nThe effective reduction in magnetic field = ",B_red," tesla "and  "\nThe change in orbit radius = ",r_change," metre ");

## Result
## The effective reduction in magnetic field = 0.132 tesla
## The change in orbit radius = 0.087 metre

The effective reduction in magnetic field =  0.13
The change in orbit radius =  0.09  metre


## Ex7-pg537¶

In [7]:
## Exa11.7 : : Page-537 (2011)
#calculate enegy of the elctron and average enegy gained per revolution
import math
B = 0.4;        ## Magnetic field, tesla
e = 1.6203e-19;        ## Charge of an electron, C
R = 30*2.54e-02;        ## Radius, metre
c = 3e+08;            ## Capacitance, farad
E = B*e*R*c/1.6e-13;        ## The energy of the electron, mega electron volts
f = 50.;                ## Frequency, cycles per sec
N = c/(4*2*math.pi*f*R);        ## Total number of revolutions
Avg_E_per_rev = E*1e+006/N;        ## Average energy gained per revolution, electron volt
print'%s %.2f %s %.2f %s '%("\nThe energy of the electron = ",E," MeV"and "\nThe average energy gained per revolution = ",Avg_E_per_rev," eV");

## Result
## The energy of the electron = 92.6 MeV
## The average energy gained per revolution = 295.57 eV
## Note: Wrong answer is given in the textbook
##   Average energy gained per revolution : 295.57 electron volts

The energy of the electron =  92.60
The average energy gained per revolution =  295.57  eV


## Ex8-pg537¶

In [6]:
## Exa11.8 : : Page-537 (2011)
#calculate peak current and average current and duty cycle
import math
R = 0.35;            ## Orbit radius, metre
N = 100e+06/480.;        ## Total number of revolutions
L = 2*math.pi*R*N;            ## Distance traversed by the electron, metre
t = 2e-06;                ## Pulse duration, sec
e = 1.6203e-19;            ## Charge of an electron, C
n = 3e+09;                ## Number of electrons
f = 180.;                ## frequency, hertz
I_p = n*e/t;            ## Peak current, ampere
I_avg = n*e*f;           ## Average current, ampere
tau = t*f;                ## Duty cycle
print'%s %.2e %s %.2e %s %.2e %s'%("\nThe peak current = ",I_p," ampere" and "\nThe average current = ",I_avg," ampere "and "\nThe duty cycle =",tau,"");

## Result
## The peak current = 2.4e-004 ampere
## The average current = 8.75e-008 ampere
## The duty cycle = 3.6e-004

The peak current =  2.43e-04
The average current =  8.75e-08
The duty cycle = 3.60e-04


## Ex9-pg538¶

In [5]:
## Exa11.9 : : Page-538 (2011)
#calculate maximum frequncy of the dee voltage and kinetic energy of the deutron
import math
q = 1.6023e-19;        ## Charge of an electron, C
B_0 = 1.5;            ## Magnetic field at the centre, tesla
m_d = 2.014102*1.66e-27;        ## Mass of the deutron, Kg
f_max = B_0*q/(2*math.pi*m_d*10**6);        ## Maximum frequency of the dee voltage, mega cycles per sec
B_prime = 1.4310;        ## Magnetic field at the periphery of the dee, tesla
f_prime = 10**7;            ## Frequency, cycles per sec
c = 3e+08;            ## Velocity of the light, metre per sec
M = B_prime*q/(2*math.pi*f_prime*1.66e-27);        ## Relativistic mass, u
K_E = (M-m_d/1.66e-27)*931.5;        ## Kinetic energy of the particle, mega electron volts
print'%s %.2f %s %.2f %s '%("\nThe maximum frequency of the dee voltage = ",f_max," MHz"and "\nThe kinetic energy of the deuteron = ",K_E," MeV");

## Result
## The maximum frequency of the dee voltage = 11.44 MHz
## The kinetic energy of the deuteron = 171.6 MeV

The maximum frequency of the dee voltage =  11.44
The kinetic energy of the deuteron =  171.62  MeV


## Ex10-pg538¶

In [4]:
#calculate frequency of the applied electric field and magnetic field intensity
## Exa11.10 : : Page-538 (2011)
import math
e = 1.6023e-19;        ## Charge of an electron, C
E = 70*1.6e-13;        ## Energy, electron volts
R = 0.28;            ## Radius of the orbit, metre
c = 3e+08;            ## Velocity of light, metre per sec
B = E/(e*R*c);        ## Magnetic field intensity, tesla
f = e*B*c**2/(2*math.pi*E);        ## Frequency, cycle per sec
del_E = 88.5*(0.07)**4*10**3/(R);     ## Energy radiated by an electron, electron volts
print'%s %.2f %s %.2f %s %.2f %s '%("\nThe frequency of the applied electric field = ",f," cycles per sec" and"\nThe magnetic field intensity = ",B," tesla"and "\nThe energy radiated by the electron =",del_E," eV");

## Result
## The frequency of the applied electric field = 1.705e+008 cycles per sec
## The magnetic field intensity = 0.832 tesla
## The energy radiated by the electron = 7.6 eV

The frequency of the applied electric field =  170523153.31
The magnetic field intensity =  0.83
The energy radiated by the electron = 7.59  eV


## Ex11-pg538¶

In [3]:
#calculate kinetic energy of the accelerated nitrogen
## Exa11.11 : : Page-538 (2011)
import math
E = 3.;        ## Energy of proton synchrotron, giga electron volts
m_0_c_sq = 0.938;        ## Relativistic energy, mega electron volts
P_p = math.sqrt(E**2-m_0_c_sq**2);        ## Momentum of the proton, giga electron volts per c
P_n = 6*P_p;        ## Momentum of the N(14) ions, giga electron volts
T_n = math.sqrt(P_n**2+(0.938*14.)**2)-0.938*14;        ## Kinetic energy of the accelerated nitrogen ion
print'%s %.2f %s'%("\nThe kinetic energy of the accelerated nitrogen ion = ",T_n," MeV");

## Result
## The kinetic energy of the accelerated nitrogen ion = 8.43 MeV

The kinetic energy of the accelerated nitrogen ion =  8.43  MeV


## Ex12-pg539¶

In [2]:
## Exa11.12 : : Page-539 (2011)
#calculate maximum magnetic flux density and maximum frequency of the accerlating voltage
import math
e = 1.6e-19;        ## Charge of an electron, C
R = 9.144;            ## Radius, metre
m_p = 1.67e-027;        ## Mass of the proton, Kg
E = 3.6*1.6e-13;        ## Energy, joule
L = 3.048;         ## Length of the one synchrotron section, metre
T = 3;            ## Kinetic energy, giga electron volts
c = 3e+08;        ## Velocity of the light, metre per sec
m_0_c_sq = 0.938;    ## Relativistic energy, mega electron volts
B = round (math.sqrt(2*m_p*E)/(R*e)*10**4);        ## Maximum magnetic field density, web per square metre
v = B*10**-4*e*R/m_p;        ## Velocity of the proton, metre per sec
f_c = v/(2*math.pi*R*10**6);        ## Frequency of the circular orbit, mega cycles per sec
f_0 = 2*math.pi*R*f_c*10**3/(2*math.pi*R+4*L);    ## Reduced frequency, kilo cycles per sec
B_m = 3.33*math.sqrt(T*(T+2*m_0_c_sq))/R;    ## Relativistic field, web per square metre
f_0 = c**2*e*R*B*1e-004/((2*math.pi*R+4*L)*(T+m_0_c_sq)*e*1e+015);    ## Maximum frequency of the accelerating voltage, mega cycles per sec
print'%s %.2f %s %.2f %s '%("\nThe maximum magnetic flux density = ",B_m," weber/Sq.m"and "\nThe maximum frequency of the accelerating voltage = ",f_0," MHz");

## Result
## The maximum magnetic flux density = 1.393 weber/Sq.m
## The maximum frequency of the accelerating voltage = 0.09 MHz
## Answer is given wrongly in the textbook

The maximum magnetic flux density =  1.39
The maximum frequency of the accelerating voltage =  0.09  MHz


## Ex13-pg539¶

In [1]:
#calculate energy of the proton
## Exa11.13 : : Page-539 (2011)
import math
E_c = 30e+009;        ## Energy of the proton accelerator, GeV
m_0_c_sq = 0.938*10**6;        ## Relativistic energy, GeV
E_p = (4*E_c**2-2*m_0_c_sq**2)/(2*m_0_c_sq) ;    ## Energy of the proton, GeV
print'%s %.2f %s'%("\nThe energy of the proton = ",E_p/1e+009," GeV");
print("wrong answer in the textbook")
## Result
## The energy of the proton = 1.92e+006 GeV
## Wrong answer given in the textbook

The energy of the proton =  1918976.54  GeV
wrong answer in the textbook