# Chapter 3 Atoms and the Bohr model¶

## Example 3.1 Page no 39¶

In [7]:
#Given
E=-3.4                                       #ev
h=6.63*10**-34                               #Js

#Calculation
import math
n=math.sqrt(-13.6/E)
M=(n*h)/(2.0*math.pi)

#Result
print"Angular momentum of electron is given by ",round(M,36)," Js"

Angular momentum of electron is given by  2.11e-34  Js


## Example 3.2 Page no 40¶

In [7]:
#Given
E=13.6                          #ev
n1=4
n2=2

#Calculation
energy=E*((1/2.0**2)-(1/4.0**2))

#Result
print"Energy of photon emitted in the transition is ",energy,"ev"

Energy of photon emitted in the transition is  2.55 ev


## Example 3.3 Page no 40¶

In [131]:
#Given
n1=3
n2=2
E1=-1.5                                       #ev
E2=-3.4                                       #ev
h=6.63*10**-34                                 #Js
c=3*10**8                                      #m/s
e=1.6*10**-19

#Calculation
v=(h*c)/((E1-E2)*e)

#Result
print"Wavelength is ",round(v,10),"m"

Wavelength is  6.543e-07 m


## Example 3.4 Page no 40¶

In [136]:
#Given
v=1200                                 #A
R=1.097*10**7                          #m-1
n1=2.0
n2=3.0

#Calculation
v1=(R*(1-(1/n1**2)))
v2=(R*(1-(1/n2**2)))
V=v1/v2
V1=V*v

#Result
print"Wavelength of the second line is ", V1,"A"

Wavelength of the second line is  1012.5 A


## Example 3.5 Page no 41¶

In [138]:
#Given
R=1.097*10**7                            #m-1
n=2

#Calculation
v=n**2/(3.0*R)
v1=1/R                                  # for n=infinite

#Result
print"longest wavelength is ",round(v*10**10,0),"A"
print"shortest wavelength is ",round(v1*10**10,1),"A"

longest wavelength is  1215.0 A
shortest wavelength is  911.6 A


## Example 3.6 Page no 41¶

In [13]:
#Given
E=47.2                               # 3ev
n1=2
n2 =3

#Calculation
import math
Z=math.sqrt(E/(13.6*((1/2.0**2)-(1/3.0**2))))

#Result
print"Atomic number of the atom is ",round(Z,0)

Atomic number of the atom is  5.0


## Example 3.7 Page no 42¶

In [143]:
#Given
Z=1.0
n=1.0                                       #for the ground state of hydrogen
Z1=4                                      #for Be++
n1=2.0

#Calculation
import math
n1=math.sqrt((n**2/Z)*Z1)
r=(Z1**2/n1**2)/(Z**2/n**2)               #Ratio of two energies

#Result
print"nBe++= ", n1
print"comparison is ",r

nBe++=  2.0
comparison is  4.0


## Example 3.8 Page no 42¶

In [144]:
#Given
Z=3.0
n=3                                          #for Li++
Z1=1.0
n1=1                                         #for hydrogen

#Calculation
r=(n**2/Z)/(n1**2/Z1)

#Result
print"orbital ratio of two states ",r

orbital ratio of two states  3.0


## Example 3.9 Page no 42¶

In [159]:
#Given
v=970.6                              #A
h=6.63*10**-34                       #Js
c=3*10**8                            #m/s

#Calculation
import math
E=((h*c)/(v*e))*10**10
En=-13.6+E
n=math.sqrt(-13.6/En)
E3=-13.6/(3.0**2)
vmax=(h*c)/((-E3+En)*(1.6*10**-19))

#Result
print"Longest wavelength is ",round(vmax*10**10),"A"

Longest wavelength is  17292.0 A


## Example 3.10 Page no 43¶

In [50]:
#Given
Z=2
E=13.6                               #ev
E0=10.04                             #ev

#Calculation
Ei=Z**2*E
E1=-Ei
E3=E1/(3.0**2)
Ee=E0+E3

#Result
print"Required stopping potential is ", round(Ee,0),"V"

Required stopping potential is  4.0 V


## Example 3.11 Page no 44¶

In [173]:
#Given
Ei=4*2.2*10**-18                           #Joule
h=6.6*10**-34                              #Js
c=3*10**8                                  #m/s

#Calculation
E1=-Ei
E2=E1/(2.0**2)
v=(h*c)/(Ei+E2)

#Result
print"Wavelength is ", round(v*10**10,0),"A"

Wavelength is  300.0 A


## Example 3.12 Page no 44¶

In [58]:
#Given
n1=3
n2 =1
E=13.6                                  #ev

#Calculation
E1=E/(3.0**2)                            #Binding energy of the atom in n=3 state
energy=E-E1                              #Energy required for the atomic electron to jump from n=1 to n=3 state

#Result
print"The electron beam must, therefore be accelerated through a potential difference of ",round(energy,2),"V"

The electron beam must, therefore be accelerated through a potential difference of  12.09 V


## Example 3.13 Page no 46¶

In [6]:
#Given
Rh=1.09678*10**7                             #m-1
Rhe=1.09722*10**7                            #m-1

#Calculation
Mr=(Rhe-Rh)/(Rh-(Rhe/4.0))                  #ratio of electron mass

#Result
print"Ratio of the electron mas to the proton mass ",round(Mr*10**4,2),"*10**-4"

Ratio of the electron mas to the proton mass  5.35 *10**-4