# Chapter 6:The Rutherford Bohr Model¶

## Example 6.1 Page 178¶

In [2]:
#initiation of variable
from math import sqrt, pi
R=0.1;Z=79.0; x=1.44;          #x=e^2/4*pi*epsi0
zkR2=2*Z*x/R                # from zkR2= (2*Z*e^2)*R^2/(4*pi*epsi0)*R^3
mv2=10.0*10**6;                #MeV=>eV

#calculation
theta=sqrt(3.0/4)*zkR2/mv2;     #deflection angle
theta=theta*(180/pi);        #converting to degrees

#result
print"Hence the average deflection angle per collision in degrees is",round(theta,3 );

Hence the average deflection angle per collision in degrees.is 0.011


## Example 6.2 Page 181¶

In [5]:
#initiation of variable
from math import sin, cos, tan, sqrt, pi
Na=6.023*10**23;p=19.3;M=197.0;
n=Na*p/M;         #The number of nuclei per atom
t=2*10**-6;Z=79;K=8*10**6;x=1.44; theta=90.0*pi/180;  #x=e^2/4*pi*epsi0
b1=t*Z*x/tan(theta/2)/(2*K)         #impact parameter b
f1=n*pi*b1**2*t                      #scattering angle greater than 90

#result
print"The fraction of alpha particles scattered at angles greater than 90 degrees is %.1e" %f1;

#part b
theta=45.0*pi/180;
b2=t*Z*x/tan(theta/2)/(2*K);
f2=n*pi*b2**2*t;                  #scattering angle greater than 45
fb=f2-f1                       #scattering angle between 45 to 90

#result
print"The fraction of particles with scattering angle from 45 to 90 is %.1e" %fb;

The fraction of alpha particles scattered at angles greater than 90 degrees is 7.5e-05
The fraction of particles with scattering angle from 45 to 90 is 3.6e-04


## Example 6.3 Page 185¶

In [7]:
#initiation of variable
from math import sin, cos, tan, sqrt, pi
Z=79.0;x=1.44;K=8.0*10**6;z=2;    #where x=e^2/4*pi*epsi0;z=2 for alpha particles

#calculation
d=z*x*Z/K;                   #distance

#result
print "The distance of closest approach in nm. is",d*10**-9

 The distance of closest approasch in nm. is 2.844e-14


## Example 6.4 Page 188¶

In [3]:
#initiation of variable
sl=820.1;n0=3.0;  #given values
n=4;w=sl*(n**2/(n**2-n0**2));
#result
print "The 3 longest possible wavelengths in nm respectively are a.",round(w,3),;

#partb
n=5.0;w=sl*(n**2/(n**2-n0**2));

#result
print "b. (in nm)",round(w,3),;

#partc
n=6.0;w=sl*(n**2/(n**2-n0**2));

#result
print "c.  (in nm )",round(w,3);

The 3 longest possible wavelengths in nm respectively are a. 1874.514 b. (in nm) 1281.406 c.  (in nm ) 1093.467


## Example 6.5 Page 189¶

In [6]:
#initiation of variable
sl=364.5;n=3.0;           #given variables and various constants are declared in the subsequent steps wherever necessary
w1=sl*(n**2/(n**2-4));     #longest wavelength of balmer
c=3.0*10**8;
f1=c/(w1*10**-9);         #corresponding freq.
n0=1.0;n=2.0;

#calculation
w2=91.13*(n**2/(n**2-n0**2)); #first longest of lymann
f2=c/(w2*10**-9);          #correspoding freq
n0=1.0;n=3.0
w3=91.13*(n**2/(n**2-n0**2));      #second longest of lymann
f3=3.0*10**8/(w3*10**-9)             #corresponding freq.

#result
print "The freq. corresponding to the longest wavelength of balmer is %.1e" %f1," & First longest wavelength  of Lymann is %.1e" %f2;
print"The sum of which s equal to %.1e" %(f1+f2);
print"The freq. corresponding to 2nd longest wavelength was found out to be %.1e" %f3,"Hence Ritz combination principle is satisfied.";

The freq. corresponding to the longest wavelength of balmer is 4.6e+14  & First longest wavelength  of Lymann is 2.5e+15
The sum of which s equal to 2.9e+15
The freq. corresponding to 2nd longest wavelength was found out to be 2.92622261239e+15 Hence Ritz combination principle is satisfied.


## Example 6.6 Page 192¶

In [9]:
#initiation of variable
Rinfi=1.097*10**7;  #known value
n1=3.0;n2=2.0;       #first 2 given states

#calculation
w=(n1**2*n2**2)/((n1**2-n2**2)*Rinfi);

#result
print"Wavelength of transition from n1=3 to n2=2 in nm is",round(w*10**9,3);

#partb
n1=4.0;n2=2.0;      #second 2 given states
w=(n1**2*n2**2)/((n1**2-n2**2)*Rinfi);

#result
print"Wavelength of transition from n1=3 to n2=2 in nm is",round(w*10**9,3);

Wavelength of trnasition from n1=3 to n2=2 in nm is 656.335
Wavelength of trnasition from n1=3 to n2=2 in nm is 486.174


## Example 6.7 Page 194¶

In [11]:
#initiation of variable
n1=3.0;n2=2.0;Z=4.0;hc=1240.0;
delE=(-13.6)*(Z**2)*((1/(n1**2))-((1/n2**2)));

#calculation
w=(hc)/delE; #for transition 1

#result
print "The wavelngth of radiation for transition(2->3) in nm is", round(w,3);

#for transition 2
n1=4.0;n2=2.0;   # n values for transition 2
delE=(-13.6)*(Z**2)*((1/n1**2)-(1/n2**2));
w=(hc)/delE;

#result
print "The wavelngth of radiation emitted for transition(2->4) in nm is", round(w,3);

The wavelngth of radiation for transition(2->3) in nm is 41.029
The wavelngth of radiation emitted for transition(2->4) in nm is 30.392