In [4]:

```
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
```

In [8]:

```
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
```

In [10]:

```
#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
```

In [15]:

```
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
```