# Chapter 3 : Elementary Quantum Physics¶

## Example 3.1 Page No : 101¶

In [1]:
# Given
lambda_val = 450*10**-9     # in nm
h = 6.6*10**-34             # in J s
e = 1.6*10**-19             # in coulombs
c = 3*10**8                 # in m/s

# Calculations and Results
E_ph = h*c/lambda_val       # in J
E_ph /= e                   # in eV
print(" Energy of blue photon in eV is {0:.4f}".format(E_ph))

 Energy of blue photon in eV is 2.7500


## Example 3.2 Page No : 103¶

In [2]:
# Given
lambda_o = 522*10**-9    # in nm
lambda_val = 250*10**-9  # in nm
h = 6.6*10**-34          # in J s
c = 3*10**8              # in m/s
e = 1.6*10**-19          # in coulombs
I = 20*10**-3            # in W/cm2
I = 20*10**-3*10**4      # in J/s/m2

# Calculations and Results
# part(a)
phi = h*c/(lambda_o*e)   # in eV
print("Work function of sodium in eV is {0:.4f}".format(phi))
KE = h*c/(lambda_val*e)-phi
print("Kinetic energy of photo emitted electrons in eV is {0:.4f}".format(KE))
J = (e*I*lambda_val)/(h*c)
print("Photoelectric current density in A/m2 is {0:.4f}".format(J))

Work function of sodium in eV is 2.3707
Kinetic energy of photo emitted electrons in eV is 2.5793
Photoelectric current density in A/m2 is 40.4040


## Example 3.4 Page No : 104¶

In [3]:
import math

# Given
theta = 15.2    # in degree
d = 0.234       # in nm
V = 100.0       # in V

# Calculations and Results
lambda_val = 2*d*math.sin(math.radians(theta))  # Braggs condition
print("Wavelength of electron in nm is {0:.4f}".format(lambda_val))
lambda_val = 1.226/math.sqrt(V)                 # debroglie wavelength in nm
print("de Broglie Wavelength of electron in nm is {0:.4f}".format(lambda_val))
print("de Broglie Wavelength is in excellent agreement with that determined from Braggs condition")

Wavelength of electron in nm is 0.1227
de Broglie Wavelength of electron in nm is 0.1226
de Broglie Wavelength is in excellent agreement with that determined from Braggs condition


## Example 3.5 Page No : 105¶

In [4]:
# Given
h = 6.6*10**-34     # in J s
c = 3*10**8         # in m/s
n = 1.0
m = 9.1*10**-31     # in Kg
a = 0.1*10**-9      # in m
e = 1.6*10**-19     # in coulombs

# Calculations and Results
E1 = (h**2*n**2)/(8*m*a**2)
E1 /= e  # in eV
print("Ground Energy of the electron in J is {0:.4f}".format(E1))
# part(b)
n = 3.0
E3 = E1*n**2
print("Energy required to put the electrons in third energy level in eV is {0:.4f}".format(E3))
E = E3-E1
print("Energy required to take the electron from E1 to E3 in eV is {0:.4f}".format(E))
lambda_val = h*c/(E*e)
print("wavelength of the required photon in nm is {0:.4g}".format(lambda_val))
print("which is an X-ray photon")

Ground Energy of the electron in J is 37.3970
Energy required to put the electrons in third energy level in eV is 336.5728
Energy required to take the electron from E1 to E3 in eV is 299.1758
wavelength of the required photon in nm is 4.136e-09
which is an X-ray photon


## Example 3.6 Page No : 110¶

In [5]:
import math

# Given
h = 6.6*10**-34     # in J s
c = 3*10**8         # in m/s
n = 1.0
m = 0.1             # in Kg
a = 1.0             # in m

# Calculations and Results
E1 = (h**2*n**2)/(8*m*a**2)
v = math.sqrt(2*E1/m)
print("Minimum speed of the object in m/s {0:.4g}".format(v))
# calculation of quantum number n
v = 1.0             # in m/s
E_n = m*v**2/2.0
n = math.sqrt((8*m*a**2*E_n)/h**2)
print("Quantum number if the object is moving with a minimum speed of 1m/s is {0:.4g}".format(n))
delta_E = (h**2/(8*m*a**2))*(2*n+1)         # delta_E=E_n+1-En
print("Separation of energy levels of the object moving with speed of 1 m/s in Joules is {0:.4g}".format(delta_E))

Minimum speed of the object in m/s 3.3e-33
Quantum number if the object is moving with a minimum speed of 1m/s is 3.03e+32
Separation of energy levels of the object moving with speed of 1 m/s in Joules is 3.3e-34


## Example 3.8 Page No : 114¶

In [6]:
# Given
h_bar = 1.054*10**-34       # in J s
delta_x = 0.1*10**-9        # in m
m_e = 9.1*10**-31           # in Kg

# Calculations and Results
delta_Px = h_bar/delta_x
print("uncertainty in momentum in Kg m/s is {0:.4g}".format(delta_Px))
delta_v = delta_Px/m_e
KE = delta_Px**2/(2*m_e)
print("Uncertainty in Kinetic Energy in J is {0:.4g}".format(KE))

uncertainty in momentum in Kg m/s is 1.054e-24
Uncertainty in Kinetic Energy in J is 6.104e-19


## Example 3.9 Page No : 115¶

In [7]:
# Given
h_bar = 1.054*10**-34       # in J s
delta_x = 1.0               # in m
m = 0.1                     # in Kg

# Calculations and Results
delta_Px = h_bar/delta_x
delta_v = delta_Px/m
print("minimum uncertainty in the velocity in m/s is {0:.4g}".format(delta_v))

minimum uncertainty in the velocity in m/s is 1.054e-33


## Example 3.10 Page No : 121¶

In [8]:
import math

# Given
h_bar = 1.054*10**-34   # in J s
m = 9.1*10**-31         # in Kg
e = 1.6*10**-19         # in coulombs
Vo = 10.0                 # in ev
Vo *= e                 # in J
E = 7.0                   # in eV
E *= e                  # in J
a = 5*10**-9            # in m

# Calculations and Results
alpha = math.sqrt(2*m*(Vo-E)/h_bar**2)
To = 16*E*(Vo-E)/Vo**2
T = To*math.exp(-2*alpha*a)
print("Transmission coefficient of conduction electrons in copper is {0:.4g}".format(T))
a = 1*10**-9  # in m
T = To*math.exp(-2*alpha*a)
print("Transmission coefficient if the oxide barrier is 1 nm is {0:.4g}".format(T))
# slight change in the answer due to approximations in alpha value

Transmission coefficient of conduction electrons in copper is 1.033e-38
Transmission coefficient if the oxide barrier is 1 nm is 6.666e-08


## Example 3.11 Page No : 122¶

In [11]:
import math

# Given
h_bar = 1.054*10**-34       # in J s
m = 100.0                     # in Kg
g = 10.0                      # in m/s2
h = 10.0                      # in m
h1 = 15.0                     # in m
a = 10.0                      # in m
E = m*g*h                   # total energy of carriage
Vo = m*g*h1                 # PE required to reach the peak

# Calculations and Results
alpha = math.sqrt(2*m*(Vo-E)/h_bar**2)
To = 16*E*(Vo-E)/Vo**2
T = To*math.exp(-2*alpha*a)
print("Transmission probability is {0:.4g}".format(T))
h_bar = 10*10**3            # Js
alpha = math.sqrt(2*m*(Vo-E)/h_bar**2)
D = Vo**2/(4*E*(Vo-E))
T = (1+(math.sinh(alpha*a))**2)**-1
print("transmission probability in a universe where h_bar is 10KJs is {0:.4g}".format(T))

Transmission probability is 0
transmission probability in a universe where h_bar is 10KJs is 0.42


## Example 3.12 Page No : 124¶

In [13]:
# Given
x = 9

# Calculation and Results
for n1 in range(1, x+1):
for n2 in range(1, x+1):
for n3 in range(1, x+1):
y = n1**2+n2**2+n3**2           # let y=N**2=n1**2+n2**2+n3**2
if y == 41:
print('{0}\t{1}\t{2}'.format(n1, n2, n3))
print("Thus there are nine possible states")

1	2	6
1	6	2
2	1	6
2	6	1
3	4	4
4	3	4
4	4	3
6	1	2
6	2	1
Thus there are nine possible states


## Example 3.13 Page No : 126¶

In [14]:
import math

# Given
h = 6.6*10**-34     # in J s
c = 3*10**8         # in m/s
m = 9.1*10**-31     # in Kg
e = 1.6*10**-19     # in coulombs
v = 2.1*10**6       # in m/s
E = m*v**2/2.0      # in J
E /= e              # in eV
E1 = -13.6          # in eV

# Calculations and Results
# change in the energy is E=En-E1
n = math.sqrt(-13.6/(E+E1))
print(" the electron gets excited to {0:.4f} level".format(n))
n = 3.0
E3 = -13.6/n**2
delta_E31 = E3-E1   # in eV
delta_E31 *= e      # in J
lambda_31 = h*c/delta_E31
print("wavelength of emitted radiation from n=3 to n=1 in nm is {0:.4f}".format(lambda_31*10**9))
# Another probability is transition fromm n=3 to n=2
n = 2.0
E2 = -13.6/n**2
delta_E32 = E3-E2  # in eV
delta_E32 *= e     # in J
lambda_32 = h*c/delta_E32
print("wavelength of emitted radiation from n=3 to n=2 in nm is {0:.4f}".format(lambda_32*10**9))
# Another probability is transition fromm n=2 to n=1
E2 = -13.6/n**2
delta_E21 = E2-E1  # in eV
delta_E21 *= e  # in J
lambda_21 = h*c/delta_E21
print("wavelength of emitted radiation from n=2 to n=1 in nm is {0:.4f}".format(lambda_21*10**9))

 the electron gets excited to 3.5835 level
wavelength of emitted radiation from n=3 to n=1 in nm is 102.3667
wavelength of emitted radiation from n=3 to n=2 in nm is 655.1471
wavelength of emitted radiation from n=2 to n=1 in nm is 121.3235


## Example 3.14 Page No : 127¶

In [15]:
# Given
Z = 2.0
n = 1.0

# Calculations and Results
E1 = -Z**2*13.6/n**2
E1 = abs(E1)
print("Energy required to ionize He+ further in eV is {0:.4f}".format(E1))

Energy required to ionize He+ further in eV is 54.4000


## Example 3.15 Page No : 130¶

In [16]:
# Given
Z = 1.0
n1 = 2.0
n2 = 3.0
R_inf = 1.0974*10**7    # in m**-1

# Calculations and Results
# Let x=1/lambda
x = R_inf*Z**2*((1/n1**2)-(1/n2**2))
lambda_val = 1.0/x
print("Wavelength of first spectral line in Angstroms is {0:.4f}".format(lambda_val*10**10))
n1 = 2.0
n2 = 4.0
x = R_inf*Z**2*((1/n1**2)-(1/n2**2))
lambda_val = 1.0/x
print("Wavelength of second spectral line in Angstroms is {0:.4f}".format(lambda_val*10**10))
print("These spectral lines correspond to H_alpha and H_beta lines of Hydrogen")

Wavelength of first spectral line in Angstroms is 6560.9623
Wavelength of second spectral line in Angstroms is 4859.9721
These spectral lines correspond to H_alpha and H_beta lines of Hydrogen


## Example 3.16 Page No : 133¶

In [17]:
# Given
h = 6.6*10**-34         # in J s
e = 1.6*10**-19         # in coulombs
E1 = 13.6               # in eV
E1 *= e                 # in J
Z = 1.0
n1 = 109.0
n2 = 110.0
ao = 52.918*10**-12     # in m

# Calculations and Results
v = Z**2*E1*((1.0/n1**2)-(1.0/n2**2))/h
print("Frequency of radiation in MHz is {0:.4f}".format(v*10**-6))
print("The frequency of radiation in the transition from n1=109 to n2=110 is same as that "
"of the detected frequency .Hence, the radiation comes from excited hydrogen atoms in"
" the give transition")
x = 2*n2**2*ao
print("The size of the atom in micro meter is {0:.4f}".format(x*10**6))
# slight difference in the answer is due to approximations

Frequency of radiation in MHz is 5022.5088
The frequency of radiation in the transition from n1=109 to n2=110 is same as that of the detected frequency .Hence, the radiation comes from excited hydrogen atoms in the give transition
The size of the atom in micro meter is 1.2806


## Example 3.20 Page No : 139¶

In [18]:
# Given
P_out = 2.5*10**-3      # in Watt
I = 5*10**-3            # in Amp
V = 2000.0              # in volts

# Calculations and Results
P_in = V*I
E = (P_out/P_in)*100
print("Efficiency of the laser in % is {0:.4f}".format(E))

Efficiency of the laser in % is 0.0250


## Example 3.21 Page No : 142¶

In [19]:
import math

# Given
lambda_o = 632.8*10**-9     # in m
c = 3*10**8                 # in m/s
T = 127.0                   # in degree celsius
T += 273.0                  # in Kelvin
m_A = 20.2*10**-3           # in Kg/mol
NA = 6.023*10**23           # mol**-1
k = 1.38*10**-23            # in J/K

# Calculations and Results
m = m_A/NA                  # in Kg
vx = math.sqrt(k*T/m)
vo = c/lambda_o
delta_v = 2*vo*vx/c
print("delta_v in GHz is {0:.4g}".format(delta_v))
delta_lambda = delta_v*(-lambda_o/vo)
print("delta_lambda in meters is {0:.4g}".format(abs(delta_lambda)))

delta_v in GHz is 1.282e+09
delta_lambda in meters is 1.711e-12