## Ex7.1:pg-200¶

In [4]:
import math
#to calculate de Broglie wavelength
v=1.5*10**7 #velocity of proton =(1/20)*velocity of light i.e.3*10**8 in m/s
m=1.67*10**-27 #mass of the proton in kg
h=6.6*10**-34 #plank's constant
lamda=h/(m*v)
print "the de Broglie wavelength is lamda=","{:.3e}".format(lamda),"m"

the de Broglie wavelength is lamda= 2.635e-14 m


## Ex7.2:pg-200¶

In [11]:
import math
#to calculate de Broglie wavelength
#mo*c**2=1.507*10**-10/1.6*10**-19=941.87 Mev
#since 12.8 Mev is very small compared to rest mass energy hence relavistic consideration may be ignored
m=1.67*10**-27 #mass in kg
h=6.62*10**-34 #plank's constant
E=12.8*10**6 #energy in Mev
lamda=h/math.sqrt(2*m*E*1.6*10**-19)/(1e-10)
print "the de Broglie wavelength is lamda=",round(lamda,5),"angstrom"

the de Broglie wavelength is lamda= 8e-05 angstrom


## Ex7.4:pg-201¶

In [6]:
import math
#to calculate wavelength
h=6.6*10**-34 #plank's constant
m=9.1*10**-31 #mass of electron in kg
E=1.25*10**3 #pottential difference keV
lamda=h/math.sqrt(2*m*E*1.6*10**-19)
print "the wavelength is lamda=","{:.2e}".format(lamda),"m"

the wavelength is lamda= 3.46e-11 m


## Ex7.5:pg-201¶

In [7]:
import math
#to calculate kinetic energy of an electron
h=6.63*10**-34 #plank's constant
mo=9.1*10**-31 #rest mass of an electron in kg
lamda=5896*10**-10  #wavelength in angstrom
K=(h**2)/(2*mo*(lamda**2)*1.6*10**-19)
print "kinetic energy of an electron is K=","{:.2e}".format(K),"eV"

kinetic energy of an electron is K= 4.34e-06 eV


## Ex7.6:pg-202¶

In [3]:
import math
#to calculate the wavelength of an electron of kinetic energy
mo=9.1*10**-31 #mass of an electron in kg
c=3*10**8 #speed of light in m/s
K=1*10**6#kinetic energy in eV
h=6.62*10**-34 #planck's constant in J-s
#E=moc**2=81.9*10**-15/1.6*10**-19 eV=0.51MeV
E=0.51*10**6
lamda=(h*c)/(math.sqrt(K*(K+2*E))*1.6*10**-19)
print "wavelength of an electron of kinetic energy is lamda=",round(lamda,16),"m"

wavelength of an electron of kinetic energy is lamda= 8.733e-13 m


## Ex7.7:pg-203¶

In [7]:
import math
#to calculate de Broglie wavelength
V=100 #potential difference in volts
lamda=12.25/math.sqrt(V)
print "de Broglie wavelength of any electron is lamda=",lamda,"angstrom"

de Broglie wavelength of any electron is lamda= 1.225 angstrom


## Ex7.9:pg-203¶

In [4]:
import math
#to calculate energy of the neutron
h=6.60*10**-34 #plank's constant in J/s
m=1.674*10**-27 #mass of the neutron in kg
lamda=10**-10 #de Broglie wavelength in m
E=(h**2)/(2*m*(lamda**2)*1.6*10**-19)
print "energy of the neutron is E=","{:.3e}".format(E),"eV"

energy of the neutron is E= 8.132e-02 eV


## Ex7.10:pg-204¶

In [8]:
import math
#to calculate wavelength
h=6.6*10**-34 #plank's constant in J/sec
m=9.1*10**-31 #mass of electron in kg
c=3*10**8 #light speed in m/s
# part a
lamda=h/(m*c)/(1e-10) # in angstrom
print "wavelength of quantum of radiant energy is lamda=",round(lamda,4),"angstrom"
#to calculate number of photons
# part b
power=12 #power emitted by the lamp =150*(8/100) in watts
E=12.0 #energy emitted per second
lamda=4500*10.0**-10
energy=(h*c)/lamda #energy contained in one photon in J
number=E/energy
print "number of photons emitted per sec is number=","{:.3e}".format(number),"unitless"

wavelength of quantum of radiant energy is lamda= 0.0242 angstrom
number of photons emitted per sec is number= 2.727e+19 unitless


## Ex7.11:pg-209¶

In [9]:
import math
#to calculate uncertainity in position
#actual formula is (delx)min*(delp)max=h/2*math.pi-------------eq(1)
#(delp)max=p(momentum of the electron)
#mv=mov/math.sqrt(1-(v/c)**2)---------------------eq(2)
mo=9*10**-31 #mass of an electron in m/s
c=3*10**8 #light speed in m/s
v=3*10**7 #velocity in m/s
h=6.6*10**-34 #plank's constant in J/s
#from eq(1) and eq(2),we get
delxmin=(h*math.sqrt(1-(v/c)**2))/(2*math.pi*mo*v)
print "smallest possible uncertainity in the position of an electron is delxmin=",round(delxmin/1e-10,3),"angstrom"

smallest possible uncertainity in the position of an electron is delxmin= 0.039 angstrom


## Ex7.12:pg-209¶

In [13]:
import math
#to calculate minimum uncertainity in the velocity
delxmax=10**-8 #maximum uncertainity in position in m
h=6.626*10**-34 #planck's constant
delpmin=h/(2*math.pi*delxmax) #minimum uncertainity in momentum in kg-m/s**2
m=9*10**-31 #mass of an electron in kg
delvmin=delpmin/m
print "minimum uncertainity in the velocity is delvmin=","{:.2e}".format(delvmin),"m/s"

minimum uncertainity in the velocity is delvmin= 1.17e+04 m/s


## Ex7.13:pg-209¶

In [10]:
import math
#to calculate uncertainity in the momentum of the parcticle
h=6.626*10**-34 #planck's constant J-s
delx=0.01*10**-2 #uncertainity in position in m
delp=h/(2*math.pi*delx)
print "uncertainity in the momentum of the parcticle is delp=","{:.2e}".format(delp),"kg-m/s**2"
#to calculate uncertainity in the velocity of an electron
m=9*10**-31 #mass of an electron in kg
delx=5*10**-10
delv=h/(2*math.pi*m*delx)
print "uncertainity in the velocity of an electron is delv=","{:.3e}".format(delv),"m/s"
#to calculate uncertainity in the velocity of alpha particle
m=4*1.67*10**-27 #mass of alpha particle in kg
delx=5*10**-10
delv=h/(2*math.pi*m*delx)
print "uncertainity in the velocity of an electron is delv=",round(delv,3),"m/s"

uncertainity in the momentum of the parcticle is delp= 1.05e-30 kg-m/s**2
uncertainity in the velocity of an electron is delv= 2.343e+05 m/s
uncertainity in the velocity of an electron is delv= 31.574 m/s


## Ex7.14:pg-210¶

In [11]:
import math
#to calculate uncertainity in position
m=9.11*10**-31 #mass of electron in kg
delv=40 #uncertainity in velocity in m/s
h=6.6*10**-34 #plank's constant
delx=h/(2*math.pi*m*delv)
print "uncertainity in the position of the electron is delx=","{:.3e}".format(delx),"m"

uncertainity in the position of the electron is delx= 2.883e-06 m


## Ex7.15:pg-210¶

In [17]:
import math
#to calculate uncertainity in frequency
#delE*delt=h/2*math.pi----eq(1)
#delE=h*delv-----------eq(2)
delt=10**-8 #uncertainity in time in s
#from eq(1) and eq(2),we get
delnu=1/(2*math.pi*delt)
print "minimum uncertainity in the frequency of the photon is delv=","{:.3e}".format(delnu),"sec**-1"

minimum uncertainity in the frequency of the photon is delv= 1.592e+07 sec**-1


## Ex7.16:pg-211¶

In [18]:
import math
#to calculate uncertainity in the energy
h=6.63*10**-34 #plank's constant in J-s
delt=2.5*10**-14 #uncertainity in time in s
delE=h/(2*math.pi*delt*1.6*10**-19)
print "minimum error with which the energy of the state can be measured is delE=",round(delE,3),"ev"

minimum error with which the energy of the state can be measured is delE= 0.026 ev


## Ex7.17:pg-211¶

In [12]:
import math
#to calculate time required for the atomic system
#delE=h*c*dellamda/lamda**2 -----eq(1)
#delE*delt=h/2*math.pi----------eq(2)
dellamda=10**-14
c=3*10**8
lamda=6*10**-7
#from eq(1)and eq(2),we get
delt=(lamda**2)/(2*math.pi*c*dellamda)
print "time required for the atomic system to retain rotational energy is delt=","{:.3e}".format(delt),"s"

time required for the atomic system to retain rotational energy is delt= 1.910e-08 s


## Ex7.18:pg-211¶

In [14]:
import math
#to calculate minimum uncertainity in the momentum
delxmax=5*10**-14 #uncertainity in position in m
h=6.626*10**-34 #plank's constant in Js
delpmin=h/(2*math.pi*delxmax)
print "minimum uncertainity in the momentum of the nucleon is delpmin=","{:.3e}".format(delpmin),"kg m/s"
m=1.675*10**-27 #mass in kg
Emin=(delpmin**2)/(2*m*1.6*10**-19)
print "minimum kinetic energy of the nucleon is Emin=",round(Emin,3),"eV"
#the answer is given wrong in the book Emin=0.039 eV

minimum uncertainity in the momentum of the nucleon is delpmin= 2.109e-21 kg m/s
minimum kinetic energy of the nucleon is Emin= 8299.24 eV


## Ex7.19:pg-212¶

In [15]:
import math
#to calculate uncertainity in velocity
delx=1.1*10**-8 #uncertainity in velocity in m
h=6.626*10**-34 #plank's constant
m=9.1*10**-31 #mass of electron in kg
delv=h/(2*math.pi*m*delx)
print "minimum uncertainity in velocity is delv=","{:.3e}".format(delv),"m/s"

minimum uncertainity in velocity is delv= 1.054e+04 m/s


## Ex7.20:pg-212¶

In [17]:
import math

#part a
#to calculate uncertainity in frequency
delt=10**-8 #uncertainity in time
delnu=1/(2*math.pi*delt)
print "minimum uncertainity in the frequency of a photon is delnu=","{:.3e}".format(delnu),"sec**-1"
#to use the uncertainity principle to place a lower limit on the energy an electron must have if it is to be part of a nucleus

# part b
delx=5*10**-15 #uncertainity in position
delp=h/(2*2*math.pi*delx) #uncertainbity in momentum
c=3*10**8 #/speed of light in m/s
E=delp*c
print "energy of an electron is E=","{:.3e}".format(E),"J"

# the answer is slightlty different due to approximation in textboook

minimum uncertainity in the frequency of a photon is delnu= 1.592e+07 sec**-1
energy of an electron is E= 3.164e-12 J


## Ex7.22:pg-223¶

In [9]:
import math
#to calculate probability of finding the particle
a=25*10**-10#width in angstrom
#wave function of the particle is chi(x)=math.sqrt(2/a)*math.sin(n*math.pi*x/a),for the particle in the least energy state n=1
chix=math.sqrt(2/a)*math.sin(math.pi*(a/2)/a)
delx=5*10**-10 #interval in angstrom
P=delx*chix**2
print "probability of finding the particle is P=",P,"unitless"

probability of finding the particle is P= 0.4 unitless


## Ex7.24:pg-224¶

In [18]:
import math
#to calculate energy of an electron
n=1 #least energy of the particle
h=6.63*10**-34 #planck's constant in Js
m=9.11*10**-31 #mass of electron in kg
a=10**-10 #width in angstrom
E=(n**2)*(h**2)/(8*m*(1.602*10**-19)*a**2)
print "energy of an electron moving in one dimension in an infinitely high potential box is E=",round(E,3),"eV"
#the answer is given wrong in the book E=5.68 eV

energy of an electron moving in one dimension in an infinitely high potential box is E= 37.649 eV


## Ex7.26:pg-225¶

In [19]:
import math
#to calculate probability
x1=0.45 #x1=0.45*L
x2=0.55       #x2=0.55*L
n=1                     #for ground state
#formula is P=integrate('(2/L)*math.sin(n*math.pi*x)**2),'x',x1,x2)
from scipy.integrate import quad
def integrand(x):
return 2*(math.sin(n*math.pi*x)**2)
P1 ,er=quad(integrand,x1,x2)

print "P1=",round(P1,3),"unitless"
probability1=P1*100
print "probability for the ground states is probability1 =",round(probability1,3),"%"
n=2   #for first excited state
P2, er=quad(integrand,x1,x2)
print "P2=",round(P2,4),"unitless"
probability2=P2*100
print "probability for first excited states is probability2=",round(probability2,3),"%"

P1= 0.198 unitless
probability for the ground states is probability1 = 19.836 %
P2= 0.0065 unitless
probability for first excited states is probability2= 0.645 %


## Ex7.28:pg-226¶

In [20]:
import math
#to calculate energy of a neutron
#consider nucleus as a cubical box of size 10**-14m
#x=y=z=a=10**-14=l
#for neutron to be in the lowest energy state nx=ny=nz=1
#formula is E=(math.pi**2*h**2/8*math.pi**2*m)*((nx/lx)**2+(ny/ly)**2+(nz/lz)**2)
h=6.626*10**-34  #planck's constant in Js
m=1.6*10**-27 #mass in kg
l=10**-14 #in m
E=(math.pi**2)*(h**2)*3/(4*(math.pi**2)*2*m*(1.6*10**-19)*l**2)
print "lowest energy of a neutron is E=",round(E/(1e6),3),"MeV"

lowest energy of a neutron is E= 6.431 MeV


## Ex7.29:pg-226¶

In [21]:
import math
#to calculate zero point energy of a linear harmonic oscillator
h=6.63*10**-34 #planck's constant in Js
nu=50 #frequency in Hz
zeropointenergy=(h*nu)/2
print "zeropointenergy=","{:.3e}".format(zeropointenergy),"J"

zeropointenergy= 1.657e-32 J


## Ex7.30:pg-226¶

In [14]:
import math
#to calculate zero point energy
nu=1 #frequency in Hz
h=6.63*10**-34 #planck's constant in Js
zeropointenergy=(h*nu)/2
print "zeropointenergy=",zeropointenergy,"J"

zeropointenergy= 3.315e-34 J


## Ex7.31:pg-226¶

In [16]:
import math
#to calculate frequency of vibration
En=0.1*1.6*10**-19 #energy of a linear harmonic oscillator in eV
n=3.0 #third excited state
h=6.63*10**-34 #planck's constant
nu=En/((n+(1/2.0))*h)
print "the frequency of vibration is nu=",round(nu,-9),"Hz"

the frequency of vibration is nu= 6.895e+12 Hz