# Chapter 1:Atomic Spectra-I¶

## Example no:1,Page no:405¶

In :
#Variable declaration
z=2                                                                             #atomic no. of He
a0=0.529                                                                        #radius of first Bohr orbit of H atom (Å)
n=1                                                                             #no. of Bohr orbit
A=2.19*10**6                                                                     #velocity of e in first Bohr orbit of H atom (m/s)
B=4.14*10**15                                                                    #orbital frequency in the first Bohr orbit of H atom (rad/s)
E0=13.6                                                                         #energy of electron in ground state of H atom (eV)
n1=1
n2=2
R=1.097*10**7                                                                    #Rydberg constant (m-1)

#Calculation
#(i) radius of first Bohr orbit
r=a0/2.0                                                                          #radius of first Bohr  orbit (Å)
#(ii) velocity of electron in the first orbit
v=A*(z/n)                                                                       #velocity of electron in the first orbit (m/s)
#(iii) orbital frequency in the first orbit
omega=B*(z**2/n**3)                                                               #orbital frequency in the first orbit (rad/s)
#(iv) kinetic and binding energy
KE=E0*(z**2/n**2)                                                                 #kinetic energy of electron in the ground state (eV)
BE=KE                                                                           #binding energy of electron in the ground state (eV)
#(v) ionization potential and first excitation potential
IP=KE                                                                           #ionization potential (eV)
EE=E0*z**2*((1.0/n1**2)-(1.0/n2**2))                                                   #first excitation potential (eV)
#(vi) wavelength of the resonance line emitted in the transition n=2 to n=1
lembda=(1.0/(R*z**2*((1.0/n1**2)-(1.0/n2**2))))*10**10                                    #wavelength of the resonance line emitted in the transition n=2 to n=1 (Å)

#Result
print"(ii) velocity =%.2e"%v,"m/s"
print"(iv) Kinetic energy =",KE,"eV    Binding energy =",BE,"eV"
print"(v) Ionization potential =",IP,"eV    EE =",EE,"eV"
print"(vi) wavelength =",lembda,"Å"

(i) radius = 0.265 Å
(ii) velocity =4.38e+06 m/s
(iii) orbital frequency = 1.656e+16 rad/s
(iv) Kinetic energy = 54.4 eV    Binding energy = 54.4 eV
(v) Ionization potential = 54.4 eV    EE = 40.8 eV
(vi) wavelength = 303.85900942 Å


## Example no:2,Page no:406¶

In :
#Variable declaration
z=1                                                                             #atomic no. of H atom
m=1.68*10**-27                                                                   #mass of H atom (kg)
h=1.054*10**-34                                                                  #Planck's constant (joule second)
R=10967800                                                                      #Rydberg constant (m-1)
e=1.6*10**-19                                                                    #Charge of electron (coulombs)
c=3*10**8                                                                        #speed of light (m/s)

#Calculation
import math
#(i) recoil velocity
v=(3*math.pi*h*R*z**2)/(2*m)                                                         #recoil velocity of H atom (m/s)
#(ii) recoil kinetic energy
Er=(9/8.0)*((math.pi*h*R*z**2)**2/(m*e))                                                #recoil kinetic energy of H atom (eV)
#(iii) energy of emitted photon
E=(1.5*math.pi*h*c*R*z**2)/e                                                         #energy of emitted photon (eV)

#Result
print"(i) recoil velocity =",round(v,2),"m/s"
print"(ii) recoil kinetic energy =%.1e"%Er,"eV"
print"(iii) energy of emitted photon =",round(E,2),"eV"

(i) recoil velocity = 3.24 m/s
(ii) recoil kinetic energy =5.5e-08 eV
(iii) energy of emitted photon = 10.21 eV


## Example no:3,Page no:407¶

In :
#Variable declaration
z=2                                                                             #atomic no. of He atom
h=1.054*10**-34                                                                  #Planck's constant (joule second)
R=10967800                                                                      #Rydberg constant (m-1)
e=1.6*10**-19                                                                    #Charge of electron (coulombs)
c=3*10**8                                                                        #speed of light (m/s)

#calculation
E=1.5*math.pi*h*c*R*z**2                                                             #The energy of the emitted photon (J)
IE=2*math.pi*h*c*R                                                                  #Ionization energy of H atom (J)
KE=(E-IE)/e                                                                     #Kinetic energy of the photoelectron (eV)

#Result
print"\nKinetic energy of photoelectron =",round(KE,1),"eV"

Kinetic energy of photoelectron = 27.2 eV


## Example no:4,Page no:407¶

In :
#Variable declaration
ratio=4                                                                         #ratio of wavelengths
z1=1                                                                            #atomic no. of hydrogen atom

#calculation
z2=math.sqrt(ratio*z1**2)                                                             #atomic no. of unknown element

#Result
print"Atomic no. =",z2,"(helium)"

Atomic no. = 2.0 (helium)


## Example no:5,Page no:407¶

In :
#Variable declaration
lembda1=108.5*10**-9                                                             #wavelength (m)
lembda2=30.4*10**-9                                                              #wavelength (m)
R=1.097*10**7                                                                    #Rydberg constant (m-1)
z=2                                                                             #atomic no. of He
n0=1                                                                            #ground state

#calculation
import math
n=math.sqrt(1.0/((1.0/n0**2)-(((1.0/lembda1)+(1.0/lembda2))/(R*z**2))))                        #quantum no. corresponding to the excited state of He+

#Result
print"n =",round(n )

n = 5.0


## Example no:6,Page no:408¶

In :
#Variable declaration
z=2                                                                             #atomic no. of He+ ion
lembda=133.7*10**-9                                                              #difference b/w the first lines of the Balmer and Lyman series (m)
n1=1
n2=2
n3=3

#calculation
R=(1.0/(lembda*z**2))*((1.0/((1.0/n2**2)-(1.0/n3**2)))-(1.0/((1.0/n1**2)-(1.0/n2**2))))            #Rudberg constant (m-1)

#Result
print"R =%.3e"%R,"m**-1"

R =1.097e+07 m**-1


## Example no:7,Page no:409¶

In :
#Variable declaration
R=1.097*10**7                                                                    #Rydberg constant (m-1)
lembda=59.3*10**-9                                                               #wavelength difference b/w first lines of Balmer and Lyman series (m)

#calculation
import math
z=math.sqrt(88.0/(15.0*R*lembda))                                                        #atomic no.

#Result
print"Z =",round(z)

Z = 3.0


## Example no:8,Page no:409¶

In :
#Variable declaration
R=1.097*10**7                                                                    #Rydberg constant (m-1)
ratio=1836                                                                      #ratio of maas of tritium and hydrogen

#calculation
lembda=(36*2*10**10)/(5*R*3*ratio)                                               #separation of the first line of the Balmer series (Å)

#Result
print"Δλ =",round(lembda,1),"Å"

Δλ = 2.4 Å