# Chapter 4:Atomic Structure¶

## Example no:4.1,Page no:125¶

In :
#Variable declaration
E= -13.6; #Energy required to separate electron and proton, eV
e= 1.6*(10**(-19)); #charge of an electron, C
E= E*e; #converting to J
Po= 8.85*(10**(-12)); #Permittivity of free space, F/m

#Calculation
import math
r= e**2/(8*(math.pi)*Po*E); #radius, m
r= -r;
m= 9.1*(10**(-31)); #mass of electron, kg
v=e/math.sqrt(4*(math.pi)*Po*m*r); #velocity, m/s

#Result
print"The orbital radius of the electron is:%.2g"%r,"m"
print"The velocity of electron is:%.2g"%v,"m/s"


The orbital radius of the electron is:5.3e-11 m
The velocity of electron is:2.2e+06 m/s


## Example no:4.2,Page no:135¶

In :
#Variable declaration
n1=1.0; #initial state
n2=3.0; #final state
E= -13.6; #energy in ground state, eV

#Calculation
dE= E*((1/n2**2)-(1/n1**2)); #Change in energy, eV

#Result
print"The energy change of Hydrogen atom is: ",round(dE,1),"eV"


The energy change of Hydrogen atom is:  12.1 eV


## Example no:4.3,Page no:135¶

In :
#Variable declaration
#Part(a)
Rn= 10.0**(-5); #radius of Rydberg atom, m
Ao= 5.29*(10**(-11)); #Bohr radius, m

#Calculation
n= math.sqrt(Rn/Ao); #Quantum number
E1= -13.6; #Ground state energy level, eV
En= E1/n**2.0; #Nth state energy level, eV

#Result
print"(a).The quantum number of the Rydberg atom is: ",round(n)
print"(b).The energy ofthe rydberg atom is:%.3g"%En,"eV"


(a).The quantum number of the Rydberg atom is:  435.0
(b).The energy ofthe rydberg atom is:-7.19e-05 eV


## Example no:4.4,Page no:138¶

In :
#Variable declaration
n1= 3.0; #initial state
n2= 2.0; #final state
R= 1.097*(10**7); #Rydberg's constant, m**(-1)

#Calculation
k= (1/n2**2)-(1/n1**2);
l= 1/(k*R); #longest wavelength, m
l= l*(10**9); #converting to nm

#Result
print"The longest in Balmer series of Hydrogen, in nm, is: ",round(l),"nm"


The longest in Balmer series of Hydrogen, in nm, is:  656.0 nm


## Example no:4.5,Page no:139¶

In :
#Variable declaration
n1=1.0; #initial state
n2=2.0; #final state
E1= 2.18*(10**(-18)); #Rydberg's constant, J
h= 6.63*(10**(-34)); #Planck's constant, J.s

#Calculation
f1= (E1/h)*(2.0/n1**3); #Frequency for first orbit, rev/s
f2= (E1/h)*(2.0/n2**3); #Frequency for second orbit, rev/s
print"Ans (A):Frequency of revolution for orbit n=1 is,f1: %.3g"%f1,"rev/s"
print"Frequency of revolution for orbit n=2 is,f2:%.2e"%f2,"rev/s"
print"which is equivalent to 0.823*10**15 rev/s 'without' any scientific notation\n"
#Part (b)
n1=2.0; #initial orbit
n2=1.0; #final orbit
f= (E1/(h))*((1.0/(n2**2))-(1.0/n1**3)); #frequency, Hz
print"Ans(B):Frequency of emitted photon is: %.3g"%f,"Hz\n"
#Part (c)
n= 2.0; #orbit
f= f2; #from part (a)
dt= 10.0**(-8); # time duration, s
N= f*dt; #Number of revolutions
#Result
print"Ans(C):Number of revolutions the electron makes is:%.3g"%N


Ans (A):Frequency of revolution for orbit n=1 is,f1: 6.58e+15 rev/s
Frequency of revolution for orbit n=2 is,f2:8.22e+14 rev/s
which is equivalent to 0.823*10**15 rev/s 'without' any scientific notation

Ans(B):Frequency of emitted photon is: 2.88e+15 Hz

Ans(C):Number of revolutions the electron makes is:8.22e+06


## Example no:4.7,Page no:142¶

In :
#Variable declaration
#Part (a)
Me= 9.1*(10**(-31)); #mass of electron, kg
m= 207*Me; #mass of muon, kg

#Calculation
Mp= 1836*Me; #mass of proton, kg
Mreduced= (m*Mp)/(m+Mp); #reduced mass, kg
Ao= 5.29*(10**(-11)); #Bohr's orbit for n=1, m
r1= Ao; #expected orbit for atom, m
r2= (Me/Mreduced)*r1; #reduced radius of orbit, m
#Part (b)
E=-13.6; # energy for elctron in n=1, eV
Ereduced= (Mreduced/Me)*E; #energy for eectron in mounic atom, eV
Ereduced= Ereduced/(10**3);#converting to keV

#Result
print"(A)Radius of the mounic atom formed, in m, is:%.3g"%r2,"m"
print"(B)Ionisation energy for the muonic atom  is: ",round(Ereduced,2),"keV"


(A)Radius of the mounic atom formed, in m, is:2.84e-13 m
(B)Ionisation energy for the muonic atom  is:  -2.53 keV


## Example no:4.8,Page no:156¶

In :
#Variable declaration
I= 7.7; #Intensity of beam, MeV
Dgold= 1.93*(10**4); #density of gold foil used, kg/m**3
u= 1.66*(10**(-27)); #atomic mass unit, kg
Mgold= 197*u; #atomic mass of gold, per atom

#Calculation
n= Dgold/Mgold; #number of atoms per unit volume, atoms/m**3
Zgold= 79; #atomic number of gold
e= 1.6*(10**(-19)); #electronis charge, C
KE= (I*e)/(10**(-6)); #converting to J
angle= 45; #degree
p=1/math.tan(math.radians(angle/2));
Po= 8.85*(10**(-12)); #Permittivity of free space, F/m
t= 3*(10**(-7)); #thickness of foil, m
f= (math.pi)*n*t*(((Zgold*(e**2))/(4*(math.pi)*Po*KE))**2)*(p**2) #using Rutherford scattering formula

#Result
print"f=%.g"%f
print"Fraction of the beam scattered through 45 degree or more is: ",round(f*100,3),"%"


f=7e-05
Fraction of the beam scattered through 45 degree or more is:  0.007 %