Atomic Models

1. Atoms

2. Thomson Model

3. Rutherford Scattering Experiment
• Bombarded Gold foil with alpha –particles ( He2+ ions )

4. Observations
• Very large number of particles went undeviated (or deviated through
small angles ) .
• Very few particles rebounded or suffered large angle deviations
• At small angles of deviations there are large
number of particles because most of the
particles suffered small angle deviations

5. Conclusions
• Atom is mostly empty
• Positive charge is concentrated in a small radius called nucleus
• Nucleus is very small as compared to Atom.
• Alpha Particle and Nuclei interaction
• Since both are positively charge they will repel each other . And
finally alpha particle will stop . Alpha particle has +2e charge
• Initial energy = (1/2)malphav2 Final Energy = K (2e) (Ze) / d

6. Bohr’s Model
• Electrons revolve around nucleus in fixed orbits having fixed energies
• For nth orbit , angular momentum about centre is :
Angular Momentum is Quantised
Whenever electron jumps from higher energy state to lower energy
state , energy is released in the form of photons .
Efinal –Einitial = hvphoton

7. Radius ,Velocity and Energy in orbit
Radius = ( It is obvious that radius will increase with n ,
but it increases with n2)
Velocity ∝
𝒁
𝒏

8. • Time period of electron in orbit
Distance = 2 pi rn ∝ n2/Z
Velocity ∝ Z/ n
Time period = Distance /Speed ∝ (n2/Z) x (n/Z) ∝ n3/Z2
Frequency = 1/T

9. Energy levels are discrete
• Total Energy = (∝ Z2/n2)
• Total energy is negative implies electron is bound
• Kinetic Energy = (KE is always +ve , ½ mv2)
• Potential Energy = Total Energy – Kinetic Energy

10. For Hydrogen
• Ground = -13.6 eV
• First excited = -3.4 eV
• Third = -1.51 eV
• Fourth = -0.85 eV
• .
• .
• Infinity = 0

11. Example : Following energies are given to H
atom .Will it accept ?
a) 10.2 eV Yes , because E2-E1 = 10.2
b) 5eV No
c) Photon of wavelength 102.5 nm
E = 1240/ 102.5 = 12.09 eV E3 – E1 =12.09
So it will accept
d) 1.89 eV Yes because E3-E2 =1.89 eV
So , If you give white light to H atom , it will accept only few
frequencies out of all .

12. • So if we give White light , it will absorb only some wavelengths .
• Also after getting excited it will emit some particular frequency
photons
• This is called atomic spectrum and is used as fingerprint for elements.
All elements have different spectrums
Only few wavelengths
absorbed 

13. Rydberg Formula
• When an electron is in excited state , it is unstable , So it jumps down
to lower states to gain some stability . When an electron jumps from
higher energy orbit to lower energy orbit , energy is emitted in the
form of photons with Wavelength

14. Name of series
• First line has smallest energy difference
 Largest λ
N=1 Lyman
N=2 Balmer
N=3 Paschen

15. Example :
• a ) 3  1
Einitial =-1.51eV Efinal= -13.6
Change = Efinal-Einitial =-13.6- (-1.51) =-12.09
-ve sign indicates energy is released .
Erelased = 12.09 eV = 1240 / λ  λ= 102.5 nm
b) Infinity  3
Einfinity =0 E3= -1.51
Change = -1.51 eV (-ve means Released )
1.51 = 1240/λ  λ =821.19 nm

16. c) Ionisation  1 – infinity
E 1 = -13.6eV Einfinity= 0
Change = final – initial = 0 – (-13.6) = + 13.6
+ve means energy is required to ionise
13.6 = 1240/ λ  λ= 91.17 nm
d) First excitation 12
Change = -3.4 –(-13.6) = +10.2eV
10.2 eV = 1240/ λ = 121.56 nm

17. A H sample is at 5 th excited state , How many
spectral lines are possible
5th excited  n=6
Possible = 65 6  4 6 3 62 6 1
54 53 52 51
43 42 41
32 31
21
Total number of lines = 15
Alternative : 6C2 = 15

18. An excited H sample is found to emit 6 different
spectral lines , find the state of sample
• Number of lines = 6
Let the sample is at nth state
nC2 = 6  n =4

19. Relation with de Broglie
• Wavelength of electron in n th orbit . Standing waves are formed ( L=n λ)
• r ∝ n2/Z  λ ∝n/Z