1) Rutherford's scattering experiment showed that the atom is mostly empty space, with the positive charge and mass concentrated in a very small nucleus at the center. 2) Bohr's model of the atom proposed that electrons orbit the nucleus in fixed, quantized energy levels. Electrons can jump between these levels, absorbing or emitting photons of specific wavelengths. 3) The Rydberg formula relates the wavelength of photons emitted during electronic transitions to the change in energy levels, allowing atomic spectra to be used to identify elements.

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