Chapter 1 Atomic Structure

Example 1_1 pgno:6

In [12]:
print'The Atomic weight of 12.0111 for Natural Carbon shows that the 12C nuclide must be present to a larger extent.'
print'\nLet 100 atoms of natural carbon contain x atoms of 12C nuclide.\n'
X=(13.0034-12.0111)*100/(13.0034-12.0000);#percentage of 12C in natural carbon#
print'Percentage of 12C in Natural carbon=X=',round(X,3)
Y=100-X;#percentage of 13C in natural carbon#
print'\nPercentage of 13C in Natural carbon=',round(Y,3)

The Atomic weight of 12.0111 for Natural Carbon shows that the 12C nuclide must be present to a larger extent.

Let 100 atoms of natural carbon contain x atoms of 12C nuclide.

Percentage of 12C in Natural carbon=X= 98.894

Percentage of 13C in Natural carbon= 1.106

Example 1_2 pgno:9

In [13]:
C=3*10**2;#velocity of light in megametre/sec#
L=300*10**-9;#wavelength of radiation in metres#
v=C*10**-6/L;#frequency of radiation in Teracycles per sec#
print'Frequency of radiation=v=Teracycles per sec=1.0*10**15Hz',v
v1=1/L;#wave number of radiation in per meter#
print'\nWavenumber of radiation=v1=3.3*10**4 per centimeter'

Frequency of radiation=v=Teracycles per sec=1.0*10**15Hz 1000.0

Wavenumber of radiation=v1=3.3*10**4 per centimeter

Example 1_3 pgno:12

In [14]:
V=0.85;#external voltage in volts#
e=1.6*10**-19;#electron charge in coloumbs#
m=9.1*10**-28;#electron mass in grams#
v=(2*V*e*10/m)**0.5;#velocity of electron in motion in Kilocm per sec#
print'velocity of electron in motion=v= Kilocm per sec=5.47*10**7cm per sec',round(v,3)
W=(3.198*10**-12)/(1.6*10**-12);#Threshold energy in eV#
print'\nThreshold energy of electron=W=eV',round(W,3)
v0=(3.198*10**-12)/(6.625*10**-15);#Threshold frequency in tera per sec#
print'\nThreshold frequency=v0=Tera per sec=4.83*10**14per sec',round(v0,3)

velocity of electron in motion=v= Kilocm per sec=5.47*10**7cm per sec 54671.848

Threshold energy of electron=W=eV 1.999

Threshold frequency=v0=Tera per sec=4.83*10**14per sec 482.717

Example 1_4 pgno:13

In [15]:
E=118.5*10**3*4.2*10**7;#energy of ions in ergs#
C=3*10**10;#velocity of light in cm/sec#
L=6.023*10**23;#Avagadro number#
h=6.625*10**-27;#plank's constant#
l=(L*h*C*10**8)/E;
print'wavelength required to cause ionization=l=Angstrums',round(l,3)

wavelength required to cause ionization=l=Angstrums 2405.206

Example 1_5 pgno:17

In [16]:
n1=2.;
n2=4.;
dE=21.7*(10**-12)*((1/n1**2)-(1/n2**2));
h=6.625*10**-27;#plank's constant#
C=3*10**10;#velocity of light in cm/sec#
l=h*C*10**8/dE;#Wavelength of second line in balmer series in Angstrums#
print'wavelength of the second line in balmer series=l=Angstrums',round(l,3)#here the answer given in textbook is slightly wrong the original answer should be the one comes through execution#

wavelength of the second line in balmer series=l=Angstrums 4884.793

Example 1_6 pgno:18

In [17]:
n1=1;
dE=21.7*(10**-12)/(1.6*10**-12*n1**2);#energy required to promote an electron from ground to infinity in eV#
print'Ionisation potential for an electron=dE=eV',round(dE,3)

Ionisation potential for an electron=dE=eV 13.563

Example 1_7 pgno:20

In [18]:
h=6.625*10**-27;#plank's constant#
V=2*10**3;#velocity of Cricket Ball in cm/sec#
m=170;#weight of Cricket Ball in grams#
l=h/(m*V);#DeBroglie Wavelength of CricketBall in Angstrums#
print'DeBroglie Wavelength of CricketBall=l==1.95*10**-24Angstrums',l

DeBroglie Wavelength of CricketBall=l==1.95*10**-24Angstrums 1.94852941176e-32

Example 1_8 pgno:21

In [19]:
from math import pi,sqrt
r1=0.53*10**-8;#Bohr radius in cm#
r2=4*r1;#Bohr radius in second state in cm#
print'Bohr radius in second state=r2=2.12*10**-8cm'
h=6.625*10**-27;#plank's constant#
m=9.11*10**-28;#electron mass in grams#
v2=h/(pi*m*r2);#electron velocity in second state in cm per sec#
print'\nElectron velocity in second state=v2=cm per sec',v2
l=(h*10**8)/(m*v2);#De Broglie wavelength of electron in second state in Angstrums#
print'\nDe Broglie wavelength of electron in second state=l=Angstrums',l
e=1.6*10**-12;#electron charge in ergs#
v=sqrt((2*(10**4)*e)/m);#velocity of the moving electron in second state in cm/sec#
print'\nVelocity of moving electron in second state=v=cm per sec',v
l1=(h*10**8)/(v*m);#De Broglie wavelength of moving elctron in Angstrums#
print'\nDe Broglie wavelength of moving electron in secondstate=l1=Angstrums',round(l1,4)

Bohr radius in second state=r2=2.12*10**-8cm

Electron velocity in second state=v2=cm per sec 109189724.953

De Broglie wavelength of electron in second state=l=Angstrums 6.66017642561

Velocity of moving electron in second state=v=cm per sec 5926738977.44

De Broglie wavelength of moving electron in secondstate=l1=Angstrums 0.1227

Example 1_9 pgno:21

In [20]:
from math import pi
m=9.11*10**-28;#electron mass in grams#
v=1.1*10**8;#velocity of electron in cm per sec#
p=m*v;#momentum of electron in gram cm per sec#
print'momentum of electron=p=10.01*10**-20gram cm per sec'
dp=p*10**-2;#Uncertainity in momentum in gram cm per sec#
print'\nUncertainity in momentum=10.01*10**-22gram cm per sec'
h=6.625*10**-27;#plank's constant#
dx=(h*10**8)/(4*pi*dp);#Uncertainity in position in Angstrum#
print'\nUncertainity in position=dx=Angstrum',round(dx,2)

momentum of electron=p=10.01*10**-20gram cm per sec

Uncertainity in momentum=10.01*10**-22gram cm per sec

Uncertainity in position=dx=Angstrum 52.61

Example 1_10 pgno:24

In [21]:
h=6.625*10**-27;#plank's constant#
g=10**3;#particle mass in grams#
l1=1.;#length of one dimensional box in cm#
n1=1.;
n2=2.;
dE1=((n2**2-n1**2)*h**2)/(8*g*l1**2);#Energy difference between two energy levels of particle in eV#
print'Energy difference between two energy levels of particle=dE1=1*10**-44eV'
l2=2*10**-8;#length of one dimensional box in cm#
m=9.11*10**-28;#electron mass in grams#
dE2=((n2**2-n1**2)*h**2)/(8*m*l2**2*1.6*10**-11);#Energy difference between two energy levels of electron in eV#
print'\nEnergy difference between two energy levels of electron=dE2=eV',round(dE2,3)
 

Energy difference between two energy levels of particle=dE1=1*10**-44eV

Energy difference between two energy levels of electron=dE2=eV 2.823