Quantum Physics

Example number 4.1, Page number 133

In [6]:
#importing modules
import math

#Variable declaration
h=6.63*10**-34;     #plancks constant in Js
m0=9.1*10**-31;     #mass of the electron in kg
c=3*10**8;       #velocity of light in m/s
phi=135;        #angle of scattering in degrees
phi=phi*0.0174532925   #converting degrees to radians  

#Calculation
delta_lamda=(h*(1-math.cos(phi)))/(m0*c);

#Result
print("change in wavelength in metres is",delta_lamda);

('change in wavelength in metres is', 4.1458307496867315e-12)

Example number 4.2, Page number 134

In [10]:
#importing modules
import math

#Variable declaration
h=6.63*10**-34;     #plancks constant in Js
m0=9.1*10**-31;     #mass of the electron in kg
c=3*10**8;       #velocity of light in m/s
lamda=2;     #wavelength in angstrom
lamdaA=lamda*10**-10;      #converting lamda from Angstrom to m
phi=90;     #angle of scattering in degrees
phi=phi*0.0174532925   #converting degrees to radians  

#Calculation
delta_lamda=(h*(1-math.cos(phi)))/(m0*c);
delta_lamda=delta_lamda*10**10;     #converting delta_lamda from m to Angstrom
delta_lamda=math.ceil(delta_lamda*10**5)/10**5;   #rounding off to 5 decimals
lamda_dash=delta_lamda+lamda;
lamdaA_dash=lamda_dash*10**-10;     #converting lamda_dash from Angstrom to m
#energy E=h*new-h*new_dash
E=h*c*((1/lamdaA)-(1/lamdaA_dash));
EeV=E/(1.602176565*10**-19);      #converting J to eV
EeV=math.ceil(EeV*10**3)/10**3;   #rounding off to 3 decimals
new=c/lamda;
new_dash=c/lamda_dash;
theta=math.atan((h*new*math.sin(phi))/((h*new)-(h*new_dash*math.cos(phi))));
theta=theta*57.2957795;       #converting radians to degrees

#Result
print("change in compton shift in Angstrom is",delta_lamda);
print("wavelength of scattered photons in Angstrom is",lamda_dash);
print("energy of recoiling electron in J is",E);
print("energy of recoiling electron in eV is",EeV);
print("angle at which recoiling electron appears in degrees is",int(theta));

#answers given in the book are wrong

('change in compton shift in Angstrom is', 0.02429)
('wavelength of scattered photons in Angstrom is', 2.02429)
('energy of recoiling electron in J is', 1.1933272900621974e-17)
('energy of recoiling electron in eV is', 74.482)
('angle at which recoiling electron appears in degrees is', 45)

Example number 4.3, Page number 135

In [13]:
#importing modules
import math

#Variable declaration
h=6.626*10**-34;     #plancks constant in Js
m0=9.1*10**-31;     #mass of the electron in kg
c=3*10**8;       #velocity of light in m/s
phi=60;     #angle of scattering in degrees
phi=phi*0.0174532925;      #converting degrees to radians
E=10**6;    #energy of photon in eV
E=E*1.6*10**-19;    #converting eV into J

#Calculation
delta_lamda=(h*(1-math.cos(phi)))/(m0*c);
delta_lamda=delta_lamda*10**10;    #converting metre to angstrom
delta_lamda=math.ceil(delta_lamda*10**4)/10**4;   #rounding off to 4 decimals
lamda=(h*c)/E;
lamdaA=lamda*10**10;     #converting metre to angstrom
lamda_dash=delta_lamda+lamdaA;
lamda_dash=math.ceil(lamda_dash*10**3)/10**3;   #rounding off to 3 decimals

#Result
print("compton shift in angstrom is",delta_lamda);
print("energy of incident photon in m",lamda);
print("wavelength of scattered photons in angstrom is",lamda_dash);

#answer for wavelength of scattered photon given in the book is wrong

('compton shift in angstrom is', 0.0122)
('energy of incident photon in m', 1.242375e-12)
('wavelength of scattered photons in angstrom is', 0.025)

Example number 4.4, Page number 135

In [15]:
#importing modules
import math

#Variable declaration
h=6.626*10**-34;     #plancks constant in Js
c=3*10**8;       #velocity of light in m/s
lamda=5893;     #wavelength in angstrom
P=60;      #output power in Watt

#Calculation
lamda=lamda*10**-10;    #wavelength in metre
E=(h*c)/lamda;
EeV=E/(1.602176565*10**-19);      #converting J to eV
EeV=math.ceil(EeV*10**4)/10**4;   #rounding off to 4 decimals
N=P/E;

#Result
print("energy of photon in J is",E);
print("energy of photon in eV is",EeV);
print("number of photons emitted per se cond is",N);

#answer for energy in eV given in the book is wrong

('energy of photon in J is', 3.373154590191753e-19)
('energy of photon in eV is', 2.1054)
('number of photons emitted per se cond is', 1.7787503773015396e+20)

Example number 4.5, Page number 136

In [18]:
#importing modules
import math

#Variable declaration
h=6.626*10**-34;     #plancks constant in Js
c=3*10**8;       #velocity of light in m/s
lamda=10;     #wavelength in angstrom

#Calculation
lamda=lamda*10**-10;    #wavelength in metre
E=(h*c)/lamda;
EeV=E/(1.602176565*10**-19);      #converting J to eV
EeV=EeV*10**-3;     #converting eV to keV
EeV=math.ceil(EeV*10**3)/10**3;   #rounding off to 3 decimals
P=h/lamda;
M=h/(lamda*c);

#Result
print("energy of photon in J is",E);
print("energy of photon in keV is",EeV);
print("momentum in kg m/sec is",P);
print("mass of photon in kg is",M);

#answer for energy of photon in keV given in the book is wrong by 1 decimal

('energy of photon in J is', 1.9878e-16)
('energy of photon in keV is', 1.241)
('momentum in kg m/sec is', 6.626e-25)
('mass of photon in kg is', 2.2086666666666664e-33)

Example number 4.6, Page number 136

In [21]:
#importing modules
import math

#Variable declaration
h=6.626*10**-34;     #plancks constant in Js
m=9.1*10**-31;     #mass of the electron in kg
e=1.602*10**-19;
V=1.25;    #potential difference in kV

#Calculation
V=V*10**3;    #converting kV to V
lamda=h/math.sqrt(2*m*e*V);
lamda=lamda*10**10;     #converting metre to angstrom
lamda=math.ceil(lamda*10**4)/10**4;   #rounding off to 4 decimals

#Result
print("de Broglie wavelength in angstrom is",lamda);

('de Broglie wavelength in angstrom is', 0.3471)

Example number 4.7, Page number 136

In [24]:
import math

#Variable declaration
E=45;    #energy of electron in eV
E=E*1.6*10**-19;     #energy in J
h=6.626*10**-34;     #plancks constant in Js
m=9.1*10**-31;     #mass of the electron in kg

#Calculation
lamda=h/math.sqrt(2*m*E);
lamda=lamda*10**10;   #converting metres to angstrom
lamda=math.ceil(lamda*10**4)/10**4;   #rounding off to 4 decimals

#Result
print("de Broglie wavelength in angstrom is",lamda);

('de Broglie wavelength in angstrom is', 1.8305)

Example number 4.8, Page number 137

In [25]:
#importing modules
import math

#Variable declaration
v=10**7;      #velocity of electron in m/sec
h=6.626*10**-34;     #plancks constant in Js
m=9.1*10**-31;     #mass of the electron in kg

#Calculation
lamda=h/(m*v);
lamda=lamda*10**10;   #converting metres to angstrom
lamda=math.ceil(lamda*10**4)/10**4;   #rounding off to 4 decimals

#Result
print("de Broglie wavelength in angstrom is",lamda);

('de Broglie wavelength in angstrom is', 0.7282)

Example number 4.9, Page number 137

In [26]:
#importing modules
import math

#Variable declaration
V=1000;       #potential difference in V
h=6.626*10**-34;     #plancks constant in Js
m=1.67*10**-27;     #mass of proton in kg
e=1.6*10**-19;      #charge of electron in J

#Calculation
lamda=h/math.sqrt(2*m*e*V);

#Result
print("de Broglie wavelength of alpha particle in metre is",lamda);

('de Broglie wavelength of alpha particle in metre is', 9.063964727801313e-13)

Example number 4.10, Page number 138

In [27]:
#importing modules
import math

#Variable declaration
L=25;    #width of potential in armstrong
delta_x=0.05;   #interval in armstrong
n=1;   #particle is in its least energy
x=L/2;   #particle is at the centre
pi=180;   #angle in degrees

#Calculation
pi=pi*0.0174532925;   #angle in radians
L=L*10**-10;   #width in m
delta_x=delta_x*10**-10;   #interval in m
#probability P = integration of (A**2)*(math.sin(n*pi*x/L))**2*delta_x
#but A=math.sqrt(2/L)
#since the particle is in a small interval integration need not be applied
#therefore P=2*(L**(-1))*(math.sin(n*pi*x/L))**2*delta_x
P=2*(L**(-1))*((math.sin(n*pi*x/L))**2)*delta_x;
P=math.ceil(P*10**3)/10**3;   #rounding off to 3 decimals

#Result
print("probability of finding the particle is",P);

('probability of finding the particle is', 0.004)

Example number 4.11, Page number 138

In [28]:
#importing modules
import math

#Variable declaration
n=1;
h=6.626*10**-34;     #plancks constant in Js
m=9.1*10**-31;     #mass of the electron in kg
L=1;    #width of potential well in angstrom

#Calculation
L=L*10**-10;     #converting angstrom into metre
E=((n**2)*h**2)/(8*m*L**2);
EeV=E/(1.6*10**-19);       #converting J to eV
EeV=math.ceil(EeV*10**3)/10**3;   #rounding off to 3 decimals

#Result
print("lowest energy of electron in J is",E);
print("lowest energy of electron in eV is",EeV);

('lowest energy of electron in J is', 6.030752197802197e-18)
('lowest energy of electron in eV is', 37.693)

Example number 4.12, Page number 139

In [29]:
#importing modules
import math

#Variable declaration
n=1;
h=6.626*10**-34;     #plancks constant in Js
m=9.1*10**-31;     #mass of the electron in kg
L=1;    #width of potential well in angstrom

#Calculation
L=L*10**-10;     #converting angstrom into metre
E=(2*(n**2)*h**2)/(8*m*L**2);
E=E/(1.6*10**-19);       #converting J to eV
E=math.ceil(E*10**3)/10**3;   #rounding off to 3 decimals

#Result
print("lowest energy of system in eV is",E);

('lowest energy of system in eV is', 75.385)

Example number 4.13, Page number 139

In [30]:
#importing modules
import math

#Variable declaration
h=6.626*10**-34;     #plancks constant in Js
m=9.1*10**-31;     #mass of the electron in kg
L=1;    #width of potential well in angstrom

#Calculation
L=L*10**-10;     #converting angstrom into metre
#according to pauli's exclusion principle, 1st electron occupies n1=1 and second electron occupies n2=2
n1=1;
n2=2;
E=((2*(n1**2)*h**2)/(8*m*L**2))+(((n2**2)*h**2)/(8*m*L**2));
E=E/(1.6*10**-19);       #converting J to eV
E=math.ceil(E*10**3)/10**3;   #rounding off to 3 decimals

#Result
print("lowest energy of system in eV is",E);
print("quantum numbers are");
print("n=1,l=0,mL=0,mS=+1/2");
print("n=1,l=0,mL=0,mS=-1/2");
print("n=2,l=0,mL=0,mS=+1/2");

('lowest energy of system in eV is', 226.154)
quantum numbers are
n=1,l=0,mL=0,mS=+1/2
n=1,l=0,mL=0,mS=-1/2
n=2,l=0,mL=0,mS=+1/2

Example number 4.14, Page number 140

In [31]:
#Variable declaration
n=1;
h=6.626*10**-34;     #plancks constant in Js
L=100;    #width of potential well in angstrom

#Calculation
L=L*10**-10;     #converting angstrom into metre
E=0.025;    #lowest energy in eV
E=E*(1.6*10**-19);       #converting eV to J
m=((n**2)*h**2)/(8*E*L**2);

#Result
print("mass of the particle in kg is",m);

('mass of the particle in kg is', 1.3719961249999998e-31)

Example number 4.15, Page number 141

In [32]:
#importing modules
import math

#Variable declaration
k=1.38*10**-23;
T=6000;      #temperature in K
h=6.626*10**-34;     #plancks constant in Js
c=3*10**8;       #velocity of light in m/s
lamda1=450;     #wavelength in nm
lamda2=460;     #wavelength in nm

#Calculation
lamda1=lamda1*10**-9;     #converting nm to metre
lamda2=lamda2*10**-9;     #converting nm to metre
new1=c/lamda1;
new2=c/lamda2;
new=(new1+new2)/2;
A=math.exp((h*new)/(k*T));
rho_v=(8*math.pi*h*new**3)/(A*c**3);

#Result
print("energy density of the black body in J/m^3 is",rho_v);

#answer given in the book is wrong

('energy density of the black body in J/m^3 is', 9.033622836188887e-16)
In [ ]: