# chapter 04 : Mechanics¶

## Ex4.1 : : Pg: 124¶

In :
from math import exp
#Percentage transmission of beam through potential barrier
eV = 1.6e-019     # Energy required by an electron to move through a potential barrier of one volt, joules
m = 9.1e-031     # Mass of electron, kg
E = 4.0*eV     # Energy of each electron, joule
Vo = 6.0*eV    # Height of potential barrier, joule
a = 10e-010     # Width of potential barrier, m
h_bar = 1.054e-34     # Reduced Planck's constant, J-s
k = 2*m*(Vo-E)/h_bar**2
# Since 2*k*a = 2*a*[2*m*(Vo-E)**1/2]/h_bar so
pow = 2.0*a/h_bar*(2*m*(Vo-E))**(1.0/2)     # Power of exponential in the expression for T
T = (16*E/Vo)*(1-E/Vo)*exp(-1*pow)     # Transmission coefficient of the beam through the potential barrier
percent_T = T*100
print "The percentage transmission of beam throught potential barrier = %5.3e %%"% percent_T

The percentage transmission of beam throught potential barrier = 1.828e-04 %


## Ex4.2 : : Pg: 125¶

In :
from math import pi
#Width of the potential barrier
A = 222.0     # Atomic weight of radioactive atom
Z = 86.0     # Atomic number of radioactive atom
eV = 1.6e-19     # Energy required by an electron to move through a potential barrier of one volt, joules
epsilon_0 = 8.854e-012     # Absolute electrical permittivity of free space, coulomb square per newton per metre square
e = 1.6e-19      # Charge on an electron, coulomb
r0 = 1.5e-015     # Nuclear radius constant, m
r = r0*A**(1.0/3)     # Radius of the radioactive atom, m
E = 4*eV*1e+06     # Kinetic energy of an alpha particle, joule
# At the distance of closest approach, r1, E = 2*(Z-2)*e**2/(4*pi*epsilon_0*r1)
# Solving for r1, we have
r1 = 2*(Z-2)*e**2/(4*pi*epsilon_0*E)     # The distance form the centre of the nucleus at which PE = KE
a = r1 - r     # Width of the potential barrier, m
print "The width of the potential barrier of the alpha particle = %5.2e m"% a

The width of the potential barrier of the alpha particle = 5.13e-14 m


## Ex4.3: : Pg : 125¶

In :
#Energy of electrons through the potential barrier
h_bar = 1.054e-34     # Reduced Planck's constant, J-s
Vo = 8e-019     # Height of potential barrier, joules
m = 9.1e-031     # Mass of an electron, kg
a = 5e-010     # Width of potential barrier, m
T = 1.0/2     # Transmission coefficient of electrons
# As T = 1/((1 + m*Vo**2*a**2)/2*E*h**2), solving for E we have
E = m*Vo**2*a**2/(2*(1/T-1)*h_bar**2*1.6e-019)     # Energy of half of the electrons through the potential barrier, eV
print "The energy of electrons through the potential barrier = %5.2f eV"% E

The energy of electrons through the potential barrier = 40.96 eV


## Ex4.4 : : Pg: 126¶

In :
from math import pi, sqrt
#Zero point energy of a system
h = 6.626e-034     # Planck's constant, Js
x = 1.0e-02     # Displacement of the spring about its mean position, m
F = 1.0e-02     # Force applied to the spring-mass system, N
m = 1.0e-03     # Mass of attached to the spring, kg
# As F = k*x, k = 4*pi**2*f**2*m is the stiffness constant, solving for f,
f = sqrt(F/(4*pi**2*m*x))     # Frequency of oscillations of mass-spring system, Hz
U = 1.0/2*h*f     # Zero point energy of the mass-spring system, J
print "The zero point energy of the mass-spring system = %4.2e J"% U

The zero point energy of the mass-spring system = 1.67e-33 J