# Ch-2 : Atomic model & bonding in solids¶

## example-2.1 page no-28¶

In :
from __future__ import division
from math import pi
#given
#atomic no. of gold
Z=79
#kinetic energy of alpha particle
E=7.68*1.6*(10)**(-13)  #J because [1MeV=1.6*(10)**(-13)]
e=1.6*10**(-19)  #C
E0=8.854*10**(-12)  #F/m
#the distance of closest approach is given by:
d0=2*e*Z*e/(4*(pi)*E0*E)  #m
print "the closest approach of alpha particle is %.2e m"%d0

the closest approach of alpha particle is 2.96e-14 m


## example-2.2 page no-29¶

In :
from math import pi, sin
#given
#IN THE RUTHERFORD SCATTERING EXPERIMENT
#the no of particles scattered at
theta1=(pi)/2  #radians
#is
N90=44  #per minute
#the number of particles scattered particales N is given by
#N=C*(1/(sin(theta/2))**4)  where C is propotionality constant
#solving above equation for C
C=N90*(sin(theta1/2))**4
# now to find the no of particles scatering at 75 and 135 degrees
theta2=75*(pi)/180  #radians
N75=C*(1/(sin(theta2/2))**4)  #per minute
theta3=135*(pi)/180  #radians
N135=C*(1/(sin(theta3/2))**4)  #per minute
print "the no of particles scattered at 75 and 135 degrees are %d per minute and %d per minutes"%(N75,N135)

the no of particles scattered at 75 and 135 degrees are 80 per minute and 15 per minutes


## example-2.3 page no-32¶

In :
#given
#mass of electron
m=9.11*10**(-31)  #kg
#charge on an electron
e=1.6*10**(-19)  #C
#plank's constant
h=6.62*10**(-34)
E0=8.85*10**(-12)
#NO OF ELECTRONS SHELLS IN HYDROZEN ATOm
n=1
#atomic number of hydrogen
Z=1
#radius of first orbit of hydrogen is given by
r1=n**2*E0*h**2/((pi)*m*Z*e**2)  #m
print "the radius of the first orbit of the electron in the hydrogen atom %.2e"%(r1)

the radius of the first orbit of the electron in the hydrogen atom 5.29e-11


## example-2.4 page no-32¶

In :
#given
#mass of electron
m=9.11*10**(-31)  #kg
#charge on an electron
e=1.6*10**(-19)  #C
#plank's constant
h=6.62*10**(-34)
E0=8.85*10**(-12)
#NO OF ELECTRONS SHELLS IN HYDROZEN ATOm
n=1
#atomic number of hydrogen
Z=1
#ionization potential energy of hydrogen atom is given by
E=m*Z**2*e**4/(8*(E0)**2*h**2*n**2)  #J
#energy in eV
EV=E/e  #eV
print "the ionization potential for hydrogen atom is %0.2f V"%(EV)

the ionization potential for hydrogen atom is 13.59 V


## example-2.5 page no-34¶

In :
#given
n=4
#the geometery of elliptical path is obtained by
#b/a=(l+1)/n
#let b/a=t just for notation
#for s suborbit
l1=0
t1=(l1+1)/n
#so b=a/4
#for p orbit
l2=1
t2=(l2+1)/n
#so b=a/2
#for d suborbit
l3=2
t3=(l3+1)/n
#so b=3*a/4
#for f suborbit
l4=3
t4=(l4+1)/n
#so b=a
print "The dimension of elliptical locus of different suborbits are :",
print t1,',',t2,',',t3,',',t4

The dimension of elliptical locus of different suborbits are : 0.25 , 0.5 , 0.75 , 1.0


## example-2.6 page no-36¶

In :
#given
#uncertainity in the momentum
deltap=10**-27  #kg ms**-1
#according to uncertainity principle
#deltap* deltax >=h/(2*(pi))
#we know that
h=6.626*10**-34  #Js
#here instead of inequality we are using only equality just for notation otherwise it is greater than equal to as mentioned above
#now deltax is given by
deltax=h/(2*(pi)*deltap)  #m
print "the minimum uncertainity is %.2e m"%(deltax)

the minimum uncertainity is 1.05e-07 m


## example-2.10 page no- 57¶

In :
#given
#ionization potential of hydrogen
E1=13.6  #eV
#when
n=3
E3=-E1/n**2  #eV
#when
n=5
E5=-E1/n**2  #eV
print "energy of 3rd and 5th orbits are %0.2f eV and %0.2f eV"%(E3,E5)

energy of 3rd and 5th orbits are -1.51 eV and -0.54 eV


## example-2.11 page no-59¶

In :
#given
#dipole moment og HF is
DM=6.375*10**(-30)  #Cm
#intermolecular distance
r=0.9178*10**(-10)  #m
#charge on an electron
e=1.67*10**(-19)  #C
#since the HF posses ionic characters
#so
#Hf in fully ionic state has dipole moment as
DM2=r*e  #Cm
#percentage ionic characters
percentage=DM/DM2*100  #%
print "the percentage ionic character is %0.2f approx."%(percentage)

the percentage ionic character is 41.59 approx.


## example-2.12 page no-60¶

In :
from math import exp
#given
#elctronegativity of In
EnIn=1.5
#elctronegativity of As
EnAs=2.2
#elctronegativity of Ga
EnGa=1.8
#for InAs
ionic_charater1=(1-exp((-0.25)*(EnAs-EnIn)**2))*100
#for GaAs
ionic_charater2=(1-exp((-0.25)*(EnAs-EnGa)**2))*100
print "percent ionic character in InAs and GaAs are %0.2f and %0.2f"%(ionic_charater1, ionic_charater2)

percent ionic character in InAs and GaAs are 11.53 and 3.92