Ch-18, Electron theories & conducting material properties, behaviour and application

example-18.1 page no-535

In [8]:
from __future__ import division
#given
#side of cube
l=20*10**-3  #m
#plank's constant
h=6.626*10**-34   #Js
#mass of electron
m=9.109*10**-31  #kg
# lower energy level
nx1=1
ny1=1
nz1=1
#higher energy level
nx2=2
ny2=1
nz2=1
#energy in the lower level
E1=h**2*(nx1**2+ny1**2+nz1**2)/(8*m*l**2)  #J
#energy in the higher level
E2=h**2*(nx2**2+ny2**2+nz2**2)/(8*m*l**2)  #J
#difference in energies of the two levels
deltaE=E2-E1  #J
print "deltaE = %0.2e"%deltaE
print "since deltaE is very small, so the assumption that E varies continuously with k, is justified"

deltaE = 4.52e-34
since deltaE is very small, so the assumption that E varies continuously with k, is justified

example-18.2 page no-537

In [3]:
#given
#distance between the plates
d=5*10**-3  #m
#voltage difference between the plates
V=230  #Volts
#electric field is given by
E=-V/d  #V/m
print "electric field between the plates is %.f V/m"%(E)

electric field between the plates is -46000 V/m

example-18.3 page no- 537

In [9]:
#given
#conductivity of material
sigma=0.018  # per ohm m
#no of electrons
n=10**19  #electrons/m**3
#voltage
V=0.16  #V
#thickness of material
t=0.29*10**-3  #m
#mass and charge of electron
m=9.1*10**-31  #kg
e=1.602*10**-19  #C
#electric feilf gradient
kie=V/t  #V/m
#as we know that
#sigma=n*e*vd/kie
#so v
vd=sigma*kie/n/e  #m/s
print " drift velocity of carriers is %0.3f m/s"%(vd)

 drift velocity of carriers is 6.199 m/s

example-18.4 page no-547

In [10]:
from math import pi
#given
#resistance of wire
R=21  #ohm
#length of wire
l=200  #m
#diameter of wire
d=0.44*10**-3  #m
#area of cross section of wire
A=(pi)/4*d**2  #m**2
#as we know that 
#R=rho*l/A
#so
rho=R*A/l  #ohm m
print "the specific resistance of wire is %0.2e ohm m"%(rho)

the specific resistance of wire is 1.60e-08 ohm m

example-18.5 page no-548

In [11]:
#given
#resistance of wire at 70 degree celcius is 57.2 ohm and at 25 degrees celcius 50 ohm
R25=50  #ohm
R70=57.2  #ohm
T1=25    #degree C
T2=70  #degree C\
#as we know that
#Rt=R0*(1+t*alpha)
#putting above values in the above given equation we get
#R25=R0*(1+25*alpha)  -------(1)
#R70=R0*(1+70*alpha)  --------(2)
#applying R70/R25
#we get
#R70/R25=(1+70*alpha)/(1+25*alpha)
#solving for alpha we get equation as follows
alpha=(R70-R25)/(T2*R25-T1*R70)  #K**-1
print "the temperature coefficient of resistance is %f K**-1"%(alpha)

the temperature coefficient of resistance is 0.003478 K**-1

example-18.6 page no-550

In [7]:
#given
#resistivity of copper , nickel and silver are
rhoCu=0.015*10**-6  #ohm m
rhoNi=0.012*10**-6  #ohm m
rhoAg=0.016*10**-6  #ohm m
#atomic percent of nickle and silver
CNi=0.25
CAg=0.40
#the resistivity of Cu-Ni-Ag alloy at 300 K
rho=rhoCu+(rhoNi*CNi)+(rhoAg*CAg)  #ohm m
print " the resistivity of Cu=Ni-Ag alloy is %.2e ohm m"%(rho)

 the resistivity of Cu=Ni-Ag alloy is 2.44e-08 ohm m