STRUCTURE OF

AT O M
CLASS OBJECTIVE

Emission and
Atomic
Recap absorption
Spectra
Spectra

Line Spectra Bohr’s Model

of Hydrogen of an atom
REVISION

Ans (i)
 REVISION
Electrons are emitted with zero velocity from a metal
surface when it is exposed to radiation of wavelength 6800
A. Calculate threshold frequency (v0) and work function
(W0) of the metal.

Ans
 REVISION
What is the number of photons of light with wavelength
4000 pm which provide 1 Joule of energy ?
 REVISION
Packet of energy is called
(a) Electron (b) Photon (c) Position
(d) Proton

Ans (b)
 REVISION
A photon of wavelength 4 × 10-7 m strikes on metal surface ;
the work function of the metal being 2.13 eV.
Calculate (i) the energy of the photon,
(ii) the kinetic energy of the emission
DUAL BEHAVIOUR OF
ELECTROMAGNETIC
RADIATION
Particle nature could explain the black body radiation and photoelectric
effect satisfactorily but on the other hand, it was not consistent with the
known wave behaviour of light which could account for the phenomena of
interference and diffraction. The only way to resolve the dilemma was to
accept the idea that light possesses both particle and wave-like
properties, i.e., light has dual behaviour.
 ATOMIC SPECTRA
▪It is observed that when a ray of white light is passed through a prism, the
wave with shorter wavelength bends more than the one with longer
wavelength.
▪Ordinary white light consists of waves with all the wavelengths in the visible
range, a ray of white light is spread out into a series of coloured bands called
spectrum.
Types of Spectrum

Emission
Absorption Spectrum
Spectrum

Continuous Line Spectra

Spectra
ATOMIC SPECTRA

Emission Spectrum :The spectrum of radiation emitted by a

substance that has absorbed energy is called an emission spectrum.

Absorption spectrum is the missing wavelength which corresponds

to the radiation absorbed by the matter, leave dark spaces in the bright
continuous spectrum
ATOMIC SPECTRA
ATOMIC SPECTRA
 LINE SPECTRUM OF
HYDROGEN
▪ When current is passed through Hydrogen gas in the discharge tube at
low pressure.

▪ The light emitted on passing electric discharge is examined with a

spectroscope.

▪ The spectrum obtained is called emission spectrum of hydrogen.

▪ It is found to consist a large number of lines, which are grouped in to

different series, named after their discoverers.
 LINE SPECTRUM OF
HYDROGEN
• Balmer showed in 1885 on the basis of experimental observations that
if spectral lines are expressed in terms of wavenumber, then the visible
lines of the hydrogen spectrum obey the following formula:

The series of lines described by this formula are called the Balmer
Series.

The Balmer series of lines are the only lines which appear in the
visible region of the electromagnetic spectrum.
LINE SPECTRUM OF
HYDROGEN
The Swedish Spectroscopist, Johannes Rydberg, noted that all series of lines in
the hydrogen spectrum could be described by the following expression:
 1. Number of lines in particular
region = n2-n1

2. Total no of spectral lines

obtained = (n2-n1) (n2 – n1 +1 ) /2

IMPORTANT 3. Ist transition = n1 = n,

FORMULAE: n2 = n +1

4. 2nd Transition-n1 = n, n2 = n +2

5. 3rd transition -n1 = n, n3= n +3

 QUESTION
• 1. Calculate the total number of spectral lines, if the electron
undergoes a transition from 7th orbit to ground state. Also calculate
the total number of spectral lines observed in Balmer region if the
transition is from 7th orbit to ground state.

(i) n2= 7, n1 =1
Total no of spectral lines = (7-1) (7-1+1)/2 = 21

(ii) n2- n1 = 7-2 =5

2. Calculate the number of spectral lines in the UV region for the

Ist transition.
Sol : n1 = 1 , n2 = 1+1
No of spectral lines = n2- n1 = 2-1 =1
NUMERICAL
• What is the wavelength of the light emitted when the
electron in a hydrogen atom undergoes transition from
the energy level with n = 4 to energy level n = 2 ? What is
the colour corresponding to this wavelength ? (Given
RH = 109678 cm-1)
 NUMERICAL
• What will be the longest wavelength in the Balmer Series of Spectra?
NUMERICAL
• Emission transitions in the Paschen series end at orbit n = 3
and start from orbit n and can be represented as v = 3.29 x
1015 (Hz) [1/32 – 1 /n2]
Calculate the value of n if the transition is observed at 1285
nm. Find the region of the spectrum.
 BOHR’S MODEL
POSTULATES
1. Electrons revolves around
the nucleus in definite
energy levels called orbits
or shells in an atom without
radiating energy. These are
numbered as 1,2,3,4…. Or
K,L,M,N
2. As long as an electron
remain in a shell it never
gains or losses energy.
 BOHR’S MODEL
POSTULATES
• 3. When energy is supplied to the electron it jumps to higher
energy level by absorbing a definite amount of energy.
• 4. The frequency of the radiation absorbed or emitted when
transition occurs between two stationary states that differ in
energy by ΔE=h ,
•  = E2 - E1 / h
 BOHR’S MODEL
POSTULATES
• 5. The angular momentum (mvr) of an electron is equal
to nh/2π.
• The angular momentum of an orbit depends upon its
quantum number
• (n) and it is integral multiple of the factor h/2π
• i.e. mvr = nh/2 π
• Where,
• n = 1, 2, 3, 4,….
 SUCCESS POINTS OF
BOHR’S MODEL
1. It could explain the stability of atom.
2. Radius of the nth orbit is given by
• rn = n2a0 / Z

• n= shell number, Z = atomic number, a0 = 52.9pm = 0.529 A0

• 3. Energy of the stationary state is given by the expression

• Joule/atom
•
Z= atomic number, n= shell number

En = -13.6 (Z2/ n2) ev/atom

En = -1312 (Z2/ n2) KJ/mol

4. Velocity of electron in different orbits is given by:

vn = 2.18x108 Z/n cm/s
 SUCCESS POINTS OF
BOHR’S MODEL
• 5. Explanation of Line Spectrum of Hydrogen
• ➢En = -2.18x10-18 Z2/ n2 J/atom
➢For Hydrogen atom Z=1
➢ ΔE = E2 –E1
➢ h  = -2.18x10-18 (1/n22 – 1/n12 )
➢ hc/ = 2.18x10-18 (1/n12 – 1/n22 )
➢ 1/ = 2.18x10-18 (1/n12 – 1/n22 )
. hc
➢ 1/ = 2.18x10-18 (1/n12 – 1/n22 )
. 6.634x10-34x3x108
➢ wave number = 109677 (1/n12 – 1/n22 ) cm-1
.

6. Applicable to one electron species – H, He+, Li2+, Be3+ etc.

 NUMERICAL
• What are the frequency and wavelength of a photon emitted during a
transition from n=5 to n=2 state in the hydrogen atom?
 NUMERICAL
• Calculate the energy and the radius of the 2 nd orbit of He+ ion.

(i) Energy = -13.6Z²/n²

= -13.6 x4/4
= -13.6 eV

(ii) Radius = 0.529 n²/Z Å

= 0.529 x4/2
= 1.058 Å
 WHAT DOES THE NEGATIVE ELECTRONIC ENERGY
OF HYDROGEN ATOM MEAN?

When the electron is at distance infinity from the nucleus, there is no

force of attraction ,the energy is zero. As the electron moves towards the
nucleus ,it experience a force of attraction . As a result some energy is
given out. Since its value was already zero, hence now it becomes negative,
Further electron comes more and more closer to the nucleus, the
attraction increases and more energy is released . Hence the energy of the
electron become more and more negative.
 1. What do you mean by saying that
energy of the electron is quantized ?
2. What is the maximum number of
emission lines obtained when the
excited electron of H-atom in n=6
HOME WORK drops to ground state. ?
3. What is the wavelength of the light
emitted when the electron in
hydrogen atom undergoes transition
from an energy level with n=4 to an
energy level with n=2?
THANK YOU STUDENTS.
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