Chapter 5 Wave packets and the uncertainty principles

Example 5.2 Page no 89

In [8]:
#Given
r=10.0**-14                                 #m
h=1.054*10**-34                             #Js
m=1.67*10**-27

#Calculation
p=h/r
E=(h**2/(2*m*(r**2)))/(1.6*10**-13)

#Result
print"Kinetic energy ",round(E,2),"Mev"

Kinetic energy  0.21 Mev

Example 5.3 Page no 89

In [44]:
#Given
E=100                                    #ev
m=9.1*10**-31
e=1.6*10**-19
x=10.0**-6                                 #m

#Calculation
import math
p=math.sqrt(2*m*E*e)
p1=h/x
theta=p1/p

#Result
print"uncertainty in the angle of emergence ", round(theta*10**4,1),"*10**-4 radians"
print "4 seconds of arc"

uncertainty in the angle of emergence  0.2 *10**-4 radians
4 seconds of arc

Example 5.4 Page no 90

In [76]:
#Given
p=0.2*10**-3*10                           #Kg m/s
h=1.054*10**-34
x=1*10**-2                                #m

#Calculation
p1=h/x
a=p1/p

#Result
print"uncertainty in the angle of emergence ",a,"radians"
print "1.1*10**-24 seconds of arc"

uncertainty in the angle of emergence  5.27e-30 radians
1.1*10**-24 seconds of arc

Example 5.5 Page no 90

In [59]:
#Given
m=50*10**-3                                    #kgram
accuracy=0.01
v=300                                    #m/s
h=1.054*10**-34

#Calculation
p=m*(v*accuracy)/100.0
x=h/p

#Result
print"position of the bullet ",round(x,32),"m"

position of the bullet  7e-32 m

Example 5.6 Page no 91

In [65]:
#Given
t=10.0**-12                                       #s
h1=1.054*10**-34
h=6.625*10**-34

#Calculation
E=h1/t
v=E/h

#Result
print"uncertainity in energy is ",E,"J"
print"uncertainity in frequency is ",round(v,-9),"Hz"

uncertainity in energy is  1.054e-22 J
uncertainity in frequency is  1.59e+11 Hz

Example 5.8 Page no 92

In [70]:
#Given
r=5*10**-15                                    #m
h=1.05*10**-34
m=1.67*10**-27
e=1.6*10**-13

#Calculation
xmax=2*r                                       #maximum  uncertainity in the position of the nucleon
pmin=h/xmax                                    #minimum uncertainity in the momentum of particle
Kmin=pmin**2/(2.0*m*e)

#Result
print"minimum kinetic energy is",round(Kmin,1),"Mev"

minimum kinetic energy is 0.2 Mev

Example 5.10 Page no 92

In [2]:
#Given
v=6000*10**-10                                #Wavelength in m
t=10**-8                              #s
c=3*10**8

#Calculation
import math
v1=v**2/(2.0*math.pi*c*t)

#Result
print"width of a line ",round(v1,15),"m"

width of a line  1.9e-14 m