# Chapter 5: Wave Properties of Matter and Quantum Mechanics I¶

## Example 5.1, Page 167¶

In [1]:
import math

#Variable declaration
N_A = 6.022e+23;    # Avogdaro's No., per mole
n = 1;    # Order of diffraction
M = 58.5;    # Molecular mass of NaCl, g/mol
rho = 2.16;    # Density of rock salt, g/cc
two_theta = 20;   # Scattering angle, degree
theta = two_theta/2;    # Diffraction angle, degree

#Calculations
N = N_A*rho*2/(M*1e-006); # Number of atoms per unit volume, per metre cube
d = (1/N)**(1./3);  # Interplanar spacing of NaCl crystal, m
lamda = 2*d*math.sin(theta*math.pi/180)/n ;    # Wavelength of X-rays using Bragg's law, m

#Result
print "The wavelength of the incident X rays  = %5.3f nm"%(lamda/1e-009)

The wavelength of the incident X rays  = 0.098 nm


## Example 5.2, Page 168¶

In [2]:
import math

#Variable declaration
h = 6.63e-034;    # Planck's constant, Js
c = 3e+008;    # Speed of light, m/s

#Calculations&Results
# For a moving ball
m = 0.057;    # Mass of the ball, kg
v = 25;    # Velocity of ball, m/s
p = m*v;    # Momentum of the ball, kgm/s
lamda = h/p;    # Lambda is the wavelength of ball, nm
print "The wavelength of ball = %3.1e m"%lamda

# For a moving electron
m = 0.511e+006;    # Rest mass of an electron, eV
K = 50;    # Kinetic energy of the electron, eV
p = math.sqrt(2*m*K);    # Momentum of the electron, kgm/s
lamda = h*c/(1.602e-019*p*1e-009);    # Wavelength of the electron, nm
print "The wavelength of the electron = %4.2f nm"%lamda

The wavelength of ball = 4.7e-34 m
The wavelength of the electron = 0.17 nm


## Example 5.3, Page 173¶

In [3]:
import math

#Variable declaration
m = 9.1e-31;    # Mass of the electron, kg
h = 6.63e-34;    # Planck's constant, Js
c = 3e+008;    # Speed of light, m/s
e = 1.6e-19;    # Energy equivalent of 1 eV, J/eV
V0 = 54;    # Potential difference between electrodes, V

#Calculations
lamda = h*c/(math.sqrt(2*m*c**2/e*V0)*e*1e-009);    # de Broglie wavelength of the electron, nm

#Result
print "The de Broglie wavelength of the electron used by Davisson and Germer = %5.3f nm"%lamda

The de Broglie wavelength of the electron used by Davisson and Germer = 0.167 nm


## Example 5.4, Page 174¶

In [4]:
import math

#Variable declaration
h = 6.63e-34;    # Planck's constant, Js
c = 3e+008;    # Speed of light, m/s
e = 1.6e-19;    # Energy equivalent of 1 eV, J/eV
m = 1.67e-27;    # Mass of a neutron, kg
k = 1.38e-23;    # Boltzmann constant, J/mol/K
T = [300, 77];    # Temperatures, K

#Calculations&Results
lamda = h*c/(math.sqrt(3*m*c**2/e*k/e*T[0])*e);    # The wavelength of the neutron at 300 K, nm
print "The wavelength of the neutron at %d K = %5.3f nm"%(T[0], lamda/1e-09)
lamda = h*c/(math.sqrt(3*m*c**2/e*k/e*T[1])*e);    # The wavelength of the neutron at 77 K, nm
print "The wavelength of the neutron at %d K = %5.3f nm"%(T[1], lamda/1e-09)

The wavelength of the neutron at 300 K = 0.146 nm
The wavelength of the neutron at 77 K = 0.287 nm


## Example 5.7, Page 184¶

In [5]:
import math

#Variable declaration
h = 6.626e-34;    # Planck's constant, Js
c = 3e+008;    # Speed of light, m/s
e = 1.602e-019;    # Energy equivalent of 1 eV, J/ev
d = 2000;    # Distance between slit centres, nm
K = 50e+003;    # Kinetic energy of electrons, eV
l = 350e+006;    # Distance of screen from the slits, nm

#Calculations
lamda = 1.226/math.sqrt(K);    # Non-relativistic value of de Broglie wavelength of the electrons, nm
E0 = 0.511e+006;    # Rest energy of the electron, J
E = K + E0;    # Total energy of the electron, J
p_c = math.sqrt(E**2 - E0**2);    # Relativistic mass energy relation, eV
lambda_r = h*c/(p_c*e*1e-009);    # Relativistic value of de Broglie wavelength, nm
percent_d = (lamda - lambda_r)/lamda*100;    # Percentage decrease in relativistic value relative to non-relavistic value
sin_theta = lambda_r/d;    # Bragg's law
y = l*sin_theta;    # The distance of first maximum from the screen, nm

#Results
print "The percentage decrease in relativistic value relative to non-relativistic value = %1.0f percent"%percent_d
print "The distance between first two maxima = %3.0f nm"%y

The percentage decrease in relativistic value relative to non-relativistic value = 2 percent
The distance between first two maxima = 938 nm


## Example 5.8, Page 187¶

In [6]:
#Variable declaration
dx = 17.5;    # The uncertainty in position, m
h = 1.05e-034;    # Reduced Planck's constant, Js

#Calculations&Results
dp_x = h/(2*dx);    # The uncertainty in momentum, kgm/s
print "The uncertainty in momentum of the ball = %1.0e kg-m/s"%dp_x
dx = 0.529e-010;    # The uncertainty in position, m
dp_x = h/(2*dx);    # The uncertainty in momentum, kgm/s
print "The uncertainty in momentum of the electron = %1.0e kg-m/s"%dp_x

The uncertainty in momentum of the ball = 3e-36 kg-m/s
The uncertainty in momentum of the electron = 1e-24 kg-m/s


## Example 5.9, Page 188¶

In [7]:
import math

#Variable declaration
a_0 = 5.29e-11;    # Radius of H-atom, m
l = 2*a_0;    # Length, m
h = 6.63e-34;    # Planck's constant, Js
m = 9.1e-31;    # Mass of electron, kg

#Calculations
K_min = h**2/(8*(math.pi)**2*m*l**2);    # Minimum kinetic energy possesed, J

#Result
print "The minimum kinetic energy of the electron = %3.1f eV"%(K_min/1.6e-19)

The minimum kinetic energy of the electron = 3.4 eV


## Example 5.10, Page 190¶

In [8]:
import math

#Variable declaration
dx = 6e-015;    # The uncertainty in position of the electron, m
h_bar = 1.054e-034;    # PReduced Planck's constant, Js
e = 1.602e-019;    # Energy equivalnet of 1 eV, J/eV
c = 3e+008;    # Speed of light, m/s
E0 = 0.511e+006;    # Rest mass energy of the electron, J

#Calculations
dp = h_bar*c/(2*dx*e);    # Minimum electron momentum, eV/c
p = dp;    # Momentum of the electron at least equal to the uncertainty in momentum, eV/c
E = math.sqrt(p**2+E0**2)/1e+006;    # Relativistic energy of the electron, MeV
K = E - E0/1e+006;    # Minimum kinetic energy of the electron, MeV

#Result
print "The minimum kinetic energy of the electron = %4.1f MeV"%K

The minimum kinetic energy of the electron = 15.9 MeV


## Example 5.12, Page 191¶

In [9]:
import math

#Variable declaration
c = 3e+8;    # Speed of light, m/s
dt = 1e-08;    # Relaxation time of atom, s
h = 6.6e-34;    # Planck's constant, Js
dE = h/(4*math.pi*dt);    # Energy width of excited state of atom, J
lamda = 300e-009;    # Wavelegth of emitted photon, m

#Calculations&Results
f = c/lamda;    # Frequency of emitted photon, per sec
print "The energy width of excited state of the atom = %3.1e eV"%(dE/1.6e-019)
df = dE/h;    # Uncertainty in frequency, per sec
print "The uncertainty ratio of the frequency = %1.0e"%(df/f)

The energy width of excited state of the atom = 3.3e-08 eV
The uncertainty ratio of the frequency = 8e-09


## Example 5.13, Page 195¶

In [10]:
#Variable declaration
e = 1.6e-019;   # Energy equivalent of 1 eV, J/eV
c = 3e008;    # Speed of light, m/s
h = 6.63e-034;    # Planck's constant, Js
m = 9.1e-031;    # Mass of the proton, kg
l = 0.1;    # Length of one-dimensional box, nm

#Calculations&Results
for n in range(1,4):
E_n = n**2*(h*c/(e*1e-009))**2/(8*m*c**2/e*l**2);    # Energy of nth level, eV
print "The first three energy levels are %3.0f eV"%(E_n)

The first three energy levels are  38 eV
The first three energy levels are 151 eV
The first three energy levels are 340 eV