Atoms Molecules and Nuclei - Part II
description
Transcript of Atoms Molecules and Nuclei - Part II
9011041155 / 9011031155
• Live Webinars (online lectures) with
recordings.
• Online Query Solving
• Online MCQ tests with detailed
solutions
• Online Notes and Solved Exercises
• Career Counseling
Atoms, Molecules and Nuclei
Origin of Spectral lines
1
9011041155 / 9011031155
If En is the energy level in nth orbit and Ep is the
energy in the pth orbit (where n > p), the energy
radiated during the fall from nth to pth orbit is,
En – Ep = hν, ---- (1) as per Bohr’s third postulate,
where ν is the frequency of radiation.But
But, according to equation (1), En – Ep = hν
But , ν = c / λ
2
9011041155 / 9011031155
Where is called Rydberg,s constant
R = 1.097 × 107 m-1
This formula gives the wavelength of the spectral line
emitted by the atom, when the excited electron jumps
from nth orbit to pth orbit.
3
9011041155 / 9011031155
Series of Spectral lines
Form the formula,
1. Lyman Series
Thus, for Lyman series, p = 1 & n = 2, 3, 4, ... and
the wavelength in U.V. range.
4
9011041155 / 9011031155
2. Balmer Series
for Balmer series, p = 2 & n = 3, 4, 5, ..& the
wavelength is given as
All the lines in this series are in visible range.
3. Paschen Series
5
9011041155 / 9011031155
Paschen series, p = 3 & n = 4, 5, 6, & the
wavelength is given as
All the lines in this series are in mid infra red range.
Two more series are emitted when the excited
electron falls to 4th and 5th orbit from higher orbits.
These series are known as Brackett series and
pfund series, respectively, which are in infra red
and far infra red regions.
6
9011041155 / 9011031155
MCQQ.11 If r1 is the first Bohr radius, then the radius of the third Bohr orbit is given by
(a.11) 3r1 (b.11) 6r1 (c.11) 9r1 (d.11)
Q.12 If the orbital velocity of the electron in the first orbit of H-atom is 2.2×106m/s, then its
orbital velocity in the second orbit is given by
(a.12) 1.1×106m/s (b.12) 4.4×106m/s (c.12) (d.12)
1.1×103m/s
Q.14 The energy required to remove an electron in a hydrogen atom from the state
corresponding to n=10 is
(a.14) 13.6eV (b.14) 1.36eV (c.14) 0.136eV (d.14) 136eV
Q.24 What is the time taken by an electron to traverse the first Bohr orbit?
(a.24) 0.50×10-16s (b.24) 1.52×10-16s (c.24) 2×10-16s (d.24) 2.52×10-16s
7
9011041155 / 9011031155
Continuous and characteristic X-rays
Coolidge tube Experiment. - It can be observed from
the nature of the graph that there is a threshold
wavelength or cutoff wavelength which is minimum,
below which no X-ray is emitted. The X-ray emitted can
be divided in two categories.
K, K wavelengths for which the intensity of X-rays is
very large. These X-rays are known as characteristic
X-rays. For remaining wavelengths intensity varies
8
9011041155 / 9011031155
gradually and corresponding X-rays are called
continuous X-rays.
The origin of continuous X-rays and cutoff wavelength
can be explained by using the relation
.Thus depends upon accelerating voltage (V) applied
and not on the material of the target on which electron
are incident.
The wavelengths for characteristic X-rays may be used
to identify the element from which they originate. For a
particular material, Wavelengths have definite values.
These X-rays emitted are called characteristics X-rays.
The value of energies are different for different
materials.
9
9011041155 / 9011031155
de Broglie’s hypothesis
1926 a French physicist de Broglie
He proposed that matter also has dual nature like light,
i.e. wave nature and particle nature. Moving particles of
matter like electrons, atoms, ions etc show wave like
properties under certain conditions. The waves
associated with moving particles are called Broglie
Waves or matter waves.
10
9011041155 / 9011031155
where h is plank’s constant
The principle is also applicable to light, when it
behaves like a photon of energy hν. If m is the mass of
the photon, moving with velocity c (velocity of light),
according to Einstein’s equation
E = mc2, mc2 = hν
this is de Broglie’s relation for photon.
1. de Broglie wavelength can be calculated only for
subatomic particles like electrons or protons.
11
9011041155 / 9011031155
2. Two velocities of the moving bodies, viz its linear
velocity and velocity of the matter wave associated
with it are different.
3. The energy carried by moving particles like
electrons is carried by virtue of its linear velocity
and not by its velocity of the matter wave.
Matter Waves
According to de Broglie, every moving particle is
associated with a wave of wavelength given by-
These waves are called matter waves.
12
9011041155 / 9011031155
As v → 0, λ → ∞ and as v → ∞, λ → 0.
Matter waves travel faster than light. The velocity of
matter wave is not constant because it depends upon
the velocity of particle. The de Broglie wavelength is
independent of the charge of particle.
The concept of matter wave is introduced
mathematically in modern quantum physics. Actually
these waves are a new kind of waves, they are not
electromagnetic in nature and are proposed to locate
the position of moving particles. The intensity of wave
at a point represents the probability of the associated
particle being there.
de Broglie’s hypothesis of matter wave also solved the
puzzle in Bohr’s 2nd postulate : why should the angular
momentum of electron have only those values that are
integral multiple of ? He thought that motion of 13
9011041155 / 9011031155
electron within atom is associated with standing wave
along the orbit as shown in Figure. We know about
standing waves in stretched strings, that only those
waves survive for which the distance travelled in round
trip between the ends is integral multiple of
wavelength.
Similarly, electron moving in nth orbit of radius rn, the
distance travelled in one trip is 2πrn that should be
integral multiple of wavelength.
By de Broglie hypothesis,
Substituting this value of ‘λ’ in above expression, 14
9011041155 / 9011031155
we get
Or
[since mvn rn is angular momentum]
angular momentum =
This is quantum condition proposed by Bohr for
angular momentum of the electron in 2nd postulate.
15
9011041155 / 9011031155
Wavelength of an electron
The de Broglie wavelength associated with an electron
can be calculated by using the relation λ = h / mv
If an electron of mass m and charge e is accelerated
by a potential difference V (volt), it will acquire energy
E given by E = eV = mv2/2
∴ m2v2 = 2meV
∴ mv = 2meV = momentum p of the electron
∴ λ = h / 2meV
Substituting the values of h.e and m we get
∴ λ = 12.27 / V
16
9011041155 / 9011031155
Davisson and Germer Experiment
1927 Davisson and Germer
17
9011041155 / 9011031155
18
9011041155 / 9011031155
Working
The beam of electrons is allowed to fall normally on the
crystal. The collector is rotated on the circular scale
and for different values of the scattering angle (θ), the
intensity (I) of the scattered electron beam is
measured, in terms of the galvanometer deflection.
Different observations are taken for different
accelerating voltages & curves showing the relation
between I and θ are plotted as follows.
19
9011041155 / 9011031155
Conclusions
A bump begins to appear in the curve for 44V electron
beam onwards. As the voltage increases, it moves
upwards, reaches a maximum value at 54V and then
decreases gradually.
For 54V electron beam, strong maxima is observed at
θ = 50o. According to de Broglie’s theory, the
wavelength associated with a 54V electron beam is
given by
This is theoretical value of wavelength λ. In Devisson
Germer experiment, the sharp diffraction pattern is due
to first maxima due to constructive interference of
electrons scattered by the crystal.
According to Bragg’s law, 2d sin φ = nλ20
9011041155 / 9011031155
In Devisson Germer experiment, φ = 65o (90 – θ/2)
and interplaner separation d for the nickel crystal is
0.91oA. For first maxima, n = 1. Substituting these
values we gat λ = 1.66oA which completely agrees
with the theoretical value. Thus, electron beam behave
as waves (X-rays because of wavelengths in that
range) and possesses wave characteristics.
• Ask Your Doubts
• For inquiry and registration, call 9011041155 /
9011031155.
21