Chapter 6 - Quantum Physics¶

Example 1 - pg 165¶

In :
#pg 165
#Quantised energy levels for microscopic and macroscopic systems
#calculate the energy levels and percentage change in energy
#Given :
import math
# (a) For a 1s simple pendulum :
T = 1.; # time period in s
nu = 1./T; #Frequency in Hz
#Planck's quantisation princple : E_n = n*h*nu
h = 6.625*10**-34 ; #Planck's constant in Js
#calculations and results
print "Energy at First three levels for a 1s simple pendulum : "
for n1 in range(1,4):
E1 = n1*h*nu ; # Energry  in J
print "E_",n1,": ",round(E1*10**34,2),"x 10^-34 J"

# (b) For a hydrogen electron
# E_n = (-13.6/n**2)eV
print "Energy at First three levels  for a hydrogen electron :"

for n2 in range(1,4):
E2 = (-13.6/n2**2);#Energy in eV
print"E_",n2,": ",round(E2,2),"eV"

#Now, for a simple pendulum
m = 10.; # mass in g
a = 1.; # amplitude in cm
omega = 2*math.pi*nu; # angular frequency in rad/s
# 1 g = 1.0*10**-3 Kg and  1 cm = 1.0*10**-2 m
E = 1./2*((m*10**-3)*(omega**2)*(a*10**-2)**2); # Energy in J
#Thus,quantum number n = E/h*nu
n = E/(h*nu);
print"Quantum number n is :",round(n*10**-28,2)," x 10**28"
#(i)Pendulum :
#percentage change in energy  = (E_n+1 - E_n)*100/E_n  which is equal to [(n+1)*h*nu - n*h*nu]*100/(n*h*nu )
#Therefore , it is (1/n) * 100
pc = (1./n)*100; #percentage change in energy
print"Percentage change in energy ( pendulum ) is",round(pc*10**27,0),"x 10**-27 "
#(ii)Hyderogen electron :
n_1 = 1; #ground state
n_2 = 2; # next quantum state
E_1 = (-13.6/n_1**2); # Energy in eV
E_2 = (-13.6/n_2**2);#Energy in eV
#percentage change : |((E_2-E_1)*100)|/ |E_1|
pc1 =((E_2-E_1)*100)/(-E_1);#percentage change
print"Percentage change in energy (hydrogen electron) is ",abs(pc1)
Energy at First three levels for a 1s simple pendulum :
E_ 1 :  6.62 x 10^-34 J
E_ 2 :  13.25 x 10^-34 J
E_ 3 :  19.88 x 10^-34 J
Energy at First three levels  for a hydrogen electron :
E_ 1 :  -13.6 eV
E_ 2 :  -3.4 eV
E_ 3 :  -1.51 eV
Quantum number n is : 2.98  x 10**28
Percentage change in energy ( pendulum ) is 3.0 x 10**-27
Percentage change in energy (hydrogen electron) is  75.0

Example 2 - pg 167¶

In :
#pg 167
#calculate the range of energies visible
#Given :
h = 6.625*10**-34;#Planck's constant in Js
c = 3*10**8 ; #velocity of light in m/s
# 1A = 1.0*10**-10 m
#(a)Energy of a photon :
# E = h*nu or E = h*c/lambd
#calculations and results
print"Energy of a photon is",round(((h*c)*10**10)*10**16,3),"x 10**-16 /lambd(in A)  J"
#1eV = 1.6*10**-19 J
print"Energy of a photon is ",round(((h*c)/(1.6*10**-19))*10**10),"/lambd(in A)  eV"
#(b)Visible light Range is 4000-7000 A
lambd1 = 4000;#Wavelength in A
lambd2 = 7000;#Wavelength in A
# 1eV = 1.6*10**-19 J ,
E1 = (h*c)/(lambd1*10**-10*1.6*10**-19); #Energy in eV
E2 = (h*c)/(lambd2*10**-10*1.6*10**-19);#Energy in eV
print"Hence the range of energies for visible photos is",round(E2,1), "eV  to",round(E1,1), "eV"
Energy of a photon is 19.875 x 10**-16 /lambd(in A)  J
Energy of a photon is  12422.0 /lambd(in A)  eV
Hence the range of energies for visible photos is 1.8 eV  to 3.1 eV

Example 3 - pg 171¶

In :
#pg 171
#calculate the take off time
import math
#Given :
#Power of the source = 10**-5 W = 10**-5 J/s
P = 10**-5 ; #Power in J/s
r = 10**-9; #radius in m
r1 = 5.; # metal plate 5 m away from the source
WF = 5.;#Work function in eV
#calculations
area = math.pi*(10**-9)**2 ; #area in m**2
area1 = 4*math.pi*r1**2; # area in m**2
P1 = P*(area/area1); # in J/s
# 1eV = 1.6*10**-19 J
t = (WF*1.6*10**-19)/P1 ;# in s
#1 day = 24 hours * 60 minutes * 60 seconds
N = t/(24*60*60); #in days
print " It will take",round(N),"days"
It will take 93.0 days

Example 4 - pg 173¶

In :
#pg 173
#calculate the Planck's constant and work function
#Given :
nu1 = 10*10**14;# Frequency in Hz
nu2 = 6*10**14;# Frequency in Hz
V_01 = 2.37; #Stopping potential in volts
V_02 = 0.72; #Stopping potential in volts
e = 1.6*10**-19 ;# Charge of an electron in C
#Einstein's photoeletric equation : h*nu = phi + e*V_0
#calculations
h = (e*(V_02 - V_01))/(nu2 - nu1);#Planck's constant in Js
phi = ((h*nu1)-(e*V_01))/e ; # work function in eV
#results
print "Plancks constant h is",h*10**34,"x 10^-34 Js  and Work function phi is",phi,"eV "
Plancks constant h is 6.6 x 10^-34 Js  and Work function phi is 1.755 eV

Example 5 - pg 176¶

In :
#pg 176
#calculate the Incident photon wavelength
#Given :
import math,numpy
ME = 35*10**3 ; #Maximum energy in eV
theta = math.pi; # photon is backscattered
h = 6.625*10**-34; #planck's constant in Js
m0 = 9.1*10**-31; #electron mass in Kg
c = 3*10**8; #Speed of light in m/s
#calculations
deltalambd = (h*(1-math.cos(theta)))/(m0*c); # in A
# (h*c/lambd) - (h*c/lambd') = 35 KeV   or  (deltalambd/lambd*lambd1) = (35 KeV/h*c)
#Simplifying the above Equation , we will obtain : lambd**2 + 0.048 lambd - 0.017
#Roots of the quadratic equation are :
values = numpy.array([1,0.048,-0.017]); # a,b,c values of the quadratic equation
r = numpy.roots(values); #Roots of the final equation
#results
print "Incident photon wavelength in Compton scattering is (A) = ",round(r,2)
Incident photon wavelength in Compton scattering is (A) =  0.11

Example 6 - pg 177¶

In :
#pg 177
#calculate the Percentage change in energy
#Given :
import math
theta = 90.; #angle in degrees
m0 = 9.1*10**-31; #electron mass in kg
c = 3.*10**8; #Speed of light in m/s
h = 6.625*10**-34; #planck's constant in Js
deltalambda = ((h*(1-math.cos(theta/57.3)))/(m0*c))*10**10; # in A
#(a) Microwave range
lambda1 = 3.0 ;# wavelength in cm
#calculations and results
#lambda1 = 3.0*10**8 A , 1 cm = 1*10**8 A
pc1 = ((deltalambda)*100.)/((lambda1*10**8) + deltalambda) ;#percent change in photon energy
print"Percentage change in energy for radiation in microwave range is :",round(pc1*10**9,0),"x 10**-9"
#(b) Visible range
lambda2 = 5000 ;# wavelength in A
pc2 = ((deltalambda)/(lambda2 + deltalambda))*100 ;#percent change in photon energy
print"Percentage change in energy for radiation in visible range is :",round(pc2*10**4,0),"x 10**-4"
#(c) X-ray range
lambda3 = 1 ; #wavelength in A
pc3 = ((deltalambda)/(lambda3 + deltalambda))*100 ;#percent change in photon energy
print"Percentage change in energy for radiation in X-ray range is :" ,round(pc3,1)
#(d)Gamma ray range
lambda4 = 0.012 ;# wavelength in A
pc4 = ((deltalambda)/(lambda4 + deltalambda))*100 ;#percent change in photon energy
print"Percentage change in energy for radiation in Gamma range is : ",round(pc4,1)
Percentage change in energy for radiation in microwave range is : 8.0 x 10**-9
Percentage change in energy for radiation in visible range is : 5.0 x 10**-4
Percentage change in energy for radiation in X-ray range is : 2.4
Percentage change in energy for radiation in Gamma range is :  66.9

Example 7 - pg 179¶

In :
#pg 179
#calculate the electron energy and momentum
#Given:
#Photoelectric effect
import math
lambda1 = 2000.; #wavelength in A
phi1 = 2.3;# Work function in eV
m = 9.1*10**-31; #electron mass in kg
E1 = 12422./lambda1; # Energy of photon in eV
c = 3.*10**8; #Speed of light in m/s
Ee1 = (12422./lambda1)- phi1; # energy of an electron in eV
pe1 = math.sqrt(2*m*Ee1*1.6*10**-19); #electron momentum in kg m/s
pp1 = (E1*1.6*10**-19)/c ; # Momentum of incident photon in kg m/s
#calculations
ratio1 = pe1/pp1 ; # (pe/pp)
#Compton effect
lambda2 = 1; # wavelength in A
deltalambda = 0.048; # Compton shift in A
E2 = 12422/lambda2; # Energy of photon in eV
Ee2 = (12422/lambda2)- (12422/(lambda2+deltalambda));#energy of an electron in eV
pe2 = math.sqrt(2*m*Ee2*1.6*10**-19); #electron momentum in kg m/s
pp2 = (E2*1.6*10**-19)/c ; # Momentum of incident photon in kg m/s
ratio2 = pe2/pp2 ; # (pe/pp)
#results
print"Photoelectric effect :";
print"Electron energy :",Ee1,"eV \n Electron momentum  :",round(pe1*10**24,2),"x 10**-24 kg m/s \n Momentum of incident photon :",round(pp1*10**27,2)," x 10**-27 kg m/s \n pe/pp :",round(ratio1,0)
print"Compton  effect :"
print"Electron energy :",round(Ee2,1),"eV \n Electron momentum  :",round(pe2*10**23,1),"x 10**-23 kg m/s \n Momentum of incident photon :",round(pp2*10**24,1),"x 10**-24 kg m/s \n pe/pp : ",round(ratio2,2)
Photoelectric effect :
Electron energy : 3.911 eV
Electron momentum  : 1.07 x 10**-24 kg m/s
Momentum of incident photon : 3.31  x 10**-27 kg m/s
pe/pp : 322.0
Compton  effect :
Electron energy : 568.9 eV
Electron momentum  : 1.3 x 10**-23 kg m/s
Momentum of incident photon : 6.6 x 10**-24 kg m/s
pe/pp :  1.94

Example 8 - pg 182¶

In :
#pg 182
#calculate the Energy, Momentum of rays
#Given:
#Gamma-rays,X-rays
lambda1 = 0.01;#Wavelength in A
c = 3*10**8; #Speed of light in m/s
#calculations
E1 =  12422./lambda1; # Energy in A
p1 = (E1*1.6*10**-19)/c ; #Momentum in kg m/s
#UV
lambda2 = 100;#Wavelength in A
c = 3*10**8; #Speed of light in m/s
E2 =  12422./lambda2; # Energy in A
p2 = (E2*1.6*10**-19)/c ; #Momentum in kg m/s
#IR
lambda3 = 1*10**-4;#Wavelength in m
c = 3*10**8; #Speed of light in m/s
#lambda3 = 1*10**-4*10**10 A , 1 m = 1*10**10 A
E3 =  12422./(lambda3*10**10); # Energy in A
p3 = (E3*1.6*10**-19)/c ; #Momentum in kg m/s
#Microwave
lambda4 = 1;#Wavelength in m
c = 3*10**8; #Speed of light in m/s
#lambda4 = 1*10**10 A , 1 m = 1*10**10 A
E4 =  12422./(lambda4*10**10); # Energy in A
p4 = (E4*1.6*10**-19)/c ; #Momentum in kg m/s
lambda5 = 100;#Wavelength in m
c = 3*10**8; #Speed of light in m/s
#lambda5 = 100*10**10 A , 1 m = 1*10**10 A
E5 =  12422./(lambda5*10**10); # Energy in A
p5 = (E5*1.6*10**-19)/c ; #Momentum in kg m/s
#results
print"Gamma-rays,X-rays : \n  Energy :",E1*10**-6," x 10**6 eV \n Momentum :",round(p1*10**22,1)," x 10**-22 kg m/s"
print" UV : \n  Energy :",E2,"eV \n Momentum :",round(p2*10**26,1),"x 10**-26 kg m/s"
print" IR : \n  Energy :",E3,"eV \n Momentum :",round(p3*10**30,1),"x 10**-30 kg m/s"
print" Microwave : \n  Energy :",E4*10**6,"x 10**-6 eV \n Momentum :",round(p4*10**34,1),"x 10**-34 kg m/s"
print" Radio waves : \n  Energy :",E5*10**8,"x 10**-8 eV \n Momentum :",round(p5*10**36,1),"x 10**-36 kg m/s"
Gamma-rays,X-rays :
Energy : 1.2422  x 10**6 eV
Momentum : 6.6  x 10**-22 kg m/s
UV :
Energy : 124.22 eV
Momentum : 6.6 x 10**-26 kg m/s
IR :
Energy : 0.012422 eV
Momentum : 6.6 x 10**-30 kg m/s
Microwave :
Energy : 1.2422 x 10**-6 eV
Momentum : 6.6 x 10**-34 kg m/s
Energy : 1.2422 x 10**-8 eV
Momentum : 6.6 x 10**-36 kg m/s

Example 9 - pg 183¶

In :
#pg 183
#calculate the wavelength
#Given :
import math
h = 6.625*10**-34; #planck's constant in Js
m = 9.109*10**-31;# electron mass in kg
e = 1.6*10**-19; # charge of an electron in C
#calculations
#lambd = h/sqrt(2*m*eV)   here we dont have V , so let us caluclate the remaining part.
lambd = h/math.sqrt(2*m*e);# wavelength in A
# 1 A = 1.0*10**-10 m
#results
print"lambd(A) =",round(lambd/(1.0*10**-10),2),"/sqrt(V) "
lambd(A) = 12.27 /sqrt(V)

Example 10 - pg 185¶

In :
#pg 185
#calculate the de Broglie wavelength for Rock and electron
#Given :
import math
#(a)Rock
h = 6.625*10**-34; #planck's constant in Js
m = 50.; # mass in g
v = 40.; # Speed in m/s
# m = 50*10**-3 kg , 1g = 1.0*10**-3 kg
#calculations
lambd1 = h/(m*10**-3*v); # Wavelength in m
#(b)For an electron
V = 50.; # in volts
lambd2 = 12.28/math.sqrt(V); # Wavelength in A
#results
print"De Broglie wavelength \n (a)Rock :",round(lambd1*10**34,2)," x 10**-34 m \n (b)For an electron (A) = ",round(lambd2,2)
De Broglie wavelength
(a)Rock : 3.31  x 10**-34 m
(b)For an electron (A) =  1.74

Example 14 - pg 193¶

In :
#pg 193
#calculate the uncertainty in position for ball and electron
#Given:
#(a) Ball
h = 6.625*10**-34; #planck's constant in Js
m1 = 45; #mass in g
v1 = 40; #Speed in m/s
prec1 = 1.5/100 ;#precision
# m1 = 45*10**-3 kg , 1 g = 1.0*10**-3 kg
#calculations and results
p1 =m1*10**-3*v1 ; # momentum in kg m/s
#(deltap/p)*100 = 1.5
deltap1 = prec1*p1 ;
deltax1 = h/deltap1; # uncertainty in position in m
print"Uncertainty in  position for a ball :",round(deltax1*10**32,2),"x 10^-32 m"
#(b) Electron
m2 = 9.1*10**-31; #electron mass in kg
v2 = 2*10**6 ; # Speed in m/s
prec2 = 1.5/100 ; # precision
p2 = m2*v2; # momentum in kg m/s
#(deltap/p)*100 = 1.5
deltap2 = prec2*p2 ;
deltax2 = h/deltap2; # uncertainty in position in m
# 1 A  = 1.0*10**-10 m
print"Uncertainty in  position for an electron (A) = ",round(deltax2/(1.0*10**-10),0)
Uncertainty in  position for a ball : 2.45 x 10^-32 m
Uncertainty in  position for an electron (A) =  243.0

Example 17 - pg 203¶

In :
#pg 203
#calculate the first 3 Energy levels of marble and electron
#Given:
#(a) Marble
h = 6.625*10**-34; #planck's constant in Js
m1 = 10.; # mass in g
L1 = 10.; # width in cm
# m1 = 10*10**-3 kg , 1 g = 1.0*10**-3 kg  and L1 = 10*10**-2 m , 1 cm = 1.0*10**-2 m
#calculations and results
print"(a)Marble "
for n1 in range (1,4):
En1 = (n1**2*h**2)/(8*m1*10**-3*(L1*10**-2)**2); # Energry  in J
print"E_",n1,": ",round(En1*10**64,1),"x 10**-64 J"

#(b) For an electron
m2 = 9.1*10**-31; #electron mass in kg
L2 = 1 ; # width in A
#L2 = 1*10**-10 m , 1 A = 1.0*10**-10 m
print"(b)For an electron "
for n2 in range(1,4):
En2 = (n2**2*h**2)/(8*m2*(L2*10**-10)**2); # Energry  in J
print"E_",n2,":",round(En2*6.24150934*10**18,1)," eV" # 1J = 6.24150934*10**18  eV
(a)Marble
E_ 1 :  5.5 x 10**-64 J
E_ 2 :  21.9 x 10**-64 J
E_ 3 :  49.4 x 10**-64 J
(b)For an electron
E_ 1 : 37.6  eV
E_ 2 : 150.5  eV
E_ 3 : 338.7  eV