# Chapter 6: Solutions of Time-Independent Schroedinger Equations¶

## Example 6.1, Page 190¶

In [1]:
import math

#Variable declaration
m=4*10**-14#in kg
v=10**-2#in m/s

#Calculations&Results
KE=(0.5*m*v**2)
h=10**-34
print "Kinetic energy(Vo-E) at %.e Joule"%KE
delta_x=(h)/math.sqrt(2*m*KE)
print "Value of penetration distance is  %.e m "%delta_x

Kinetic energy(Vo-E) at 2e-18 Joule
Value of penetration distance is  2e-19 m


## Example 6.2, Page 191¶

In [2]:
import math

#Variable declaration
#KE=4ev, convert to joule
KE=4*1.6*10**-19#in j
m=9*10**-31#in kg
h=10**-34#in j-s

#Calculations
delta_x=(h)/math.sqrt(2*m*KE)

#Result
print "Value of penetration distance is  %.1e m "%delta_x

Value of penetration distance is  9.3e-11 m


## Example 6.3, Page 198¶

In [3]:
import math

#Variable declaration
v=50.#in Mev
E=55.#in Mev

#Calculations
x=math.sqrt(1-(v/E))
R=((1-x)/(1+x))**2

#Result
print "Probablity of neutron will be reflected is  %.2f"%R

Probablity of neutron will be reflected is  0.29


## Example 6.4, Page 202¶

In [5]:
#Variable declaration
m=9*10**-31#in kg
h=10**-34#in j-s
V=10#in ev
a=1.8*10**-10#in m

#Calculations
#convert v to joule
Vo=V*1.6*10**-19#in Joule
N=(2*m*Vo*a**2)/(h**2)

#Result
print "Number given is  %d "%N

Number given is  9


## Example 6.6, Page 219¶

In [6]:
import math

#Variable declaration
h=10**-34#in j-s
m=10**-30#in kg
a=10**-14#in m
c=3*10**8#in m/s

#Calculations&Results
E=((math.pi*h)**2)/(2*m*a*a)
print "Energy is %.e  J"%E
#convert to ev
e=E/(1.6*10**-19)
print "Energy is %.e  eV "%e
#Answer difference is due to round off
E1=(math.pi*c*h)/a
print "Zero level Energy is %.e  J "%E1
e1=E1/(1.6*10**-19)
print "Zero level Energy is %.e  ev "%e1
#Answer difference is due to round off
#when A=100
A=100
r=10**-14#in m
x=10**-10#in coul2/nt-m2
ec=1.6*10**-19#in c
Q=(-(A*ec*ec)/(x*r))*(1/ec)
print "Typical value Energy is %.e  ev "%Q

Energy is 5e-10  J
Energy is 3e+09  eV
Zero level Energy is 9e-12  J
Zero level Energy is 6e+07  ev
Typical value Energy is -2e+07  ev