# Chapter 6: Quantum Mechanics II¶

## Example 6.4, Page 205¶

In :
import scipy
import math

#Variable declaration
alpha = 1;    # For simplicity assume alpha to be unity

#Calculations&Results
p = lambda x: math.sqrt(alpha)*math.exp(-2*alpha*x)    # Probability that the particle lies between 0 and 1/alpha
print "The probability that the particle lies between 0 and 1/alpha = %5.3f"%integ
p = lambda x: math.sqrt(alpha)*math.exp(-2*alpha*x)
integ, err = scipy.integrate.quad(p,1/alpha,2/alpha) # Probability that the particle lies between 1/alpha and 2/alpha
print "The probability that the particle lies between 1/alpha and 2/alpha = %5.3f"%integ

The probability that the particle lies between 0 and 1/alpha = 0.432
The probability that the particle lies between 1/alpha and 2/alpha = 0.059


## Example 6.9, Page 215¶

In :
import math

#Variable declaration
h = 6.62e-034;    # Planck's constant, Js
h_bar = h/(2*math.pi);    # Reduced Planck's constant, Js
c = 3.00e+008;    # Speed of light, m/s
e = 1.602e-019;    # Energy equivalent of 1 eV, J
m = 938.3e+006;    # Energy equivalent of proton mass, eV
L = 1e-005;    # Diameter of the nucleus, nm

#Calculations
E1 = math.pi**2*(h_bar*c/(e*1e-009))**2/(2*L**2*m*1e+006);      # Energy of the ground state of proton, MeV
E2 = 4*E1;    # Energy of first excited state of proton, MeV
delta_E = E2 -E1;    # Transition energy of the proton inside the nucleus, MeV

#Result
print "The transition energy of the proton inside the nucleus = %1d MeV"%delta_E

The transition energy of the proton inside the nucleus = 6 MeV


## Example 6.14, Page 229¶

In :
import math

#Variable declaration
h = 6.62e-034;    # Planck's constant, Js
h_bar = h/(2*math.pi);    # Reduced Planck's constant, Js
c = 3.00e+008;    # Speed of light, m/s
e = 1.602e-019;    # Energy equivalent of 1 eV, J
m = 0.511e+006;    # Energy equivalent of electron rest mass, eV
V0 = 10;    # Height of potential barrier, eV
E = 5;    # Energy of the incident electrons, eV
L = 0.8e-009;    # Width of the potential barrier, m

#Calculations
k = math.sqrt(2*m*(V0 - E))*e/(h_bar*c);    # Schrodinger's constant, per m
T = (1 + V0**2*math.sinh(k*L)**2/(4*E*(V0 - E)))**(-1);    # Transmission electron probability

#Result
print "The fraction of electrons tunnelled through the barrier = %3.1e"%T

The fraction of electrons tunnelled through the barrier = 4.4e-08


## Example 6.15, Page 229¶

In :
import math

#Variable declaration
h = 6.62e-034;    # Planck's constant, Js
h_bar = h/(2*math.pi);    # Reduced Planck's constant, Js
c = 3.00e+008;    # Speed of light, m/s
e = 1.602e-019;    # Energy equivalent of 1 eV, J
m = 0.511e+006;    # Energy equivalent of electron rest mass, eV
V0 = 10;    # Height of potential barrier, eV

#Calculations
Sum_M = 0;
i = 1;
for E in range(1,5):    # Range of energies of the incident electrons, eV
M = 16*E/V0*(1-E/V0);    # All the factors multiplying the exponential term
Sum_M = Sum_M + M;    # Accumulator
i = i + 1;

E = 5;    # Given energy of the incident electrons, eV
M = int(Sum_M/i);    # Average value of M
L = 0.8e-009;    # Width of the potential barrier, m
k = math.sqrt(2*m*(V0 - E))*e/(h_bar*c);    # Schrodinger's constant, per m
T = M*math.exp(-2*k*L);    # Transmission electron probability

#Result
print "The fraction of electrons tunnelled through the barrier = %3.1e"%T

The fraction of electrons tunnelled through the barrier = 2.2e-08


## Example 6.16, Page 230¶

In :
import math

#Variable declaration
h = 6.62e-034;    # Planck's constant, Js
h_bar = h/(2*math.pi);    # Reduced Planck's constant, Js
c = 3.00e+008;    # Speed of light, m/s
e = 1.602e-019;    # Energy equivalent of 1 eV, J
m = 0.511e+006;    # Energy equivalent of electron rest mass, eV
V0 = 10;    # Height of potential barrier, eV
E = 5;    # Given energy of the incident electrons, eV

#Calculations
l = h_bar*c/(2*math.sqrt(2*m*(V0 - E))*1e-009*e);    # Penetration distance into the barrier when the probability of the particle penetration drops to 1/e, nm

#Result
print "The penetration distance for a %d eV electron approaching a step barrier of %d eV = %5.3f nm"%(E, V0, l)

The penetration distance for a 5 eV electron approaching a step barrier of 10 eV = 0.044 nm


## Example 6.17, Page 234¶

In :
import math

#Variable declaration
h = 6.62e-034;    # Planck's constant, Js
h_bar = h/(2*math.pi);    # Reduced Planck's constant, Js
c = 3.00e+008;    # Speed of light, m/s
e = 1.602e-019;    # Charge on an electron, C
k = 9e+009;    # Coulomb constant, N-Sq.m./C^2
m = 3727;    # Energy equivalent of alpha particle rest mass, MeV
E = 5;    # Given energy of the incident electrons, eV
Z1 = 2;    # Atomic number of an alpha particle
Z2 = 92;    # Atomic number of the U-238 nucleus
r_N = 7e-015;    # Nuclear radius, m
K = 4.2;    # Kinetic energy of alpha particle, MeV

#Calculations
V_C = Z1*Z2*e**2*k/(r_N*e*1e+006);    # Coulomb Potential, MeV
r_prime = V_C*r_N/(K*1e-015);    # Distance through which the alpha particle must tunnel, fm
kapa = math.sqrt(2*m*(V_C-E))*e/(h_bar*c*1e-006);    # Schronginger's Constant, per m
L = r_prime - r_N/1e-015;    # Barrier width, fm
T = 16*(K/V_C)*(1-K/V_C)*math.exp(-2*kapa*L*1e-015);    # Tunnelling probability of alpha particle
V_C_new = V_C/2;    # Potential equal to half the Coulomb potential, MeV
L = L/2;    # Width equal to half the barrier width, fm
kapa = math.sqrt(2*m*(V_C_new-E))*e/(h_bar*c*1e-006);    # Schronginger's Constant, per m
T_a = 16*(K/V_C_new)*(1-K/V_C_new)*math.exp(-2*kapa*L*1e-015);    # Approximated tunnelling probability of alpha particle
v = math.sqrt(2*K/m)*c;    # Speed of the alpha particle, m/s
t = r_N/v;    # Time taken by alpha particle to cross the U-238 nucleus, s

#Results
print "The barrier height = %2d MeV"%(math.ceil(V_C))
print "The distance that alpha particle must tunnel = %2d fm"%r_prime
print "The tunnelling potential assuming square-top potential = %1.0e"%T
print "The approximated tunnelling potential = %3.1e"%T_a
print "The speed of the alpha particle = %3.1e m/s"%v
print "The time taken by alpha particle to cross the U-238 nucleus = %1.0e s"%t

# Some answers are given wrong in the textbook for this problem

The barrier height = 38 MeV
The distance that alpha particle must tunnel = 63 fm
The tunnelling potential assuming square-top potential = 6e-123
The approximated tunnelling potential = 3.8e-40
The speed of the alpha particle = 1.4e+07 m/s
The time taken by alpha particle to cross the U-238 nucleus = 5e-22 s