Atoms
Notes On: Atomic Nucleus, Dual Nature.
Distance of closest approach (a0): When 𝛼 particles are bombarded over a target nucleus it approaches to a
closest separation before momentarily coming to rest. This happens when kinetic energy of 𝛼 particles is
converted into electrostatic potential energy. This distance is called distance of closest approach.
1
2
mu2 =
𝑎0 =
1
𝑍𝑒 (2𝑒)
4𝜋𝜀0
𝑎0
1
Target
X-(He++)
1
2𝑍𝑒 2
2
4𝜋𝜀0 1𝑚𝑢2
2
mu2
Ze
a0
Nucleus
Impact Parameter: The collision of a projectile with a target nucleus can be of two types
Head on collision
& oblique collision. The factor Which determines whether a collision is oblique or head on is impact parameter.
It is defined as the Perpendicular distance
1
He
++
m𝑢2
2
𝜃
b/w the line passing from the
b
centre of the target nucleus and || to the
velocity vector of 𝛼- particle and the velocity vector itself. (see fig.)
It is given as
b=
1
𝑍𝑒 2 𝑐𝑜𝑡𝜃/2
4𝜋𝜀0
1
𝑚𝑢2
2
b
Impact parameter
Ze
Target Nucleus
For head-on collision b= 0
For oblique collision b≠ 0
Marsden Experiment (In support with Rutherford atomic model
This was the first experiment to conceptualize structure of atom. A theoretical model was proposed by
Rutherford but its experimental verification was done by his two research scholar Geiger and Marsden.
Exp. Set up:
𝛼 − 𝑃𝑜𝑟𝑡𝑖𝑐𝑙𝑒
Gold Foil
Experimental results:
(i)
Most –𝛼 – particles went past the thin foil straight or with slight deviation.
(ii)
A few deflected with angles > 900
(iii)
Abt ‘1’ in 8000 retrace der path.
(iv)
1
The no. of ‘𝛼’ particles ‘N’ diffracted by angle ∅ is givan as 𝑆𝑖𝑛4 (∅/2) ∝ LogeN.
N
Graph b/w No. of ′𝛼’ particles
defected at angle ‘ 𝜃‘ vs ‘𝜃’
00
900
1800
𝜃
Bohr atomic model: - (For H & H – like atom)
Postulates: (1) Atom consists of a solid part at its center called nucleus which consists of Protons and
neutrons.
(2)
As long as ‘e’ is revolving in its orbit its’ energy and angular momentum remains conserved.
Thus the orbits are called stationary orbits.
(3)
The angular momentum ofrevolving electrons is integral multiple of ‘h/2𝜋′ ie.
mvr =
(4)
𝑛ℎ
2𝜋
When ‘e’ Jumps from one orbit to another, energy is absorbed or evolved depending upon
the kind of Jump (Lower to higher or higher to lower energy levels)
Expressions of radius, speed and total energy of an electron in its orbit :
vn
Force on the electron in nth orbit
e
2e
rn
rn
F=
2
𝑚 𝑣𝑛
𝑟𝑛
m 𝑣𝑛2 =
Also:
mvnrn =
1
=
4𝜋𝜀0 𝑟𝑛2
𝑘 𝑍𝑒 2
𝑟𝑛
--------------------------------
(1)
𝑛ℎ
2𝜋
𝑛ℎ
vn = 2𝜋𝑚𝑟
𝑛
From eqn (1) & (2):
𝑍𝑒.𝑒
----------------------- (2)
𝑛2 ℎ 2
m.4𝜋2 𝑚2 𝑟 2 =
𝑛
rn =
𝑘𝑍𝑒 2
𝑟𝑛
𝑛2 ℎ 2
--------------------------------
4𝜋2 𝑚𝑘𝑍𝑒 2
ℎ2
r = 4𝜋2 𝑚𝑘𝑒 2 ≈ 0.529A0
For H atom:
From (3) & (2):
𝑛ℎ ×4𝜋2 𝑚𝑘𝑍𝑒 2
vn =
2𝜋𝑚.𝑛2 ℎ2
Vn =
2𝜋𝑘𝑧𝑒 2
𝑛ℎ
for H-atom, V ≈ 1.8 x 106 ms-1
Now Total energy: En = K.E + P.E
1
= 2mv2 –
En = -
𝑘𝑧𝑒 2
𝑟𝑛
1
{ As T.E = 2(PE)
2𝑟𝑛
En = −
𝑘.𝑧𝑒 2
4𝜋 2 𝑚𝑧 2 𝑒 4 𝑘 2
𝑛2 ℎ 2
For H-atom,
= −
13.6 𝑒𝑣 2
𝑧
𝑛2
En = −
13.6 𝑒𝑣
𝑛2
Spectrum of H-atom: (Absorption Spectrum)
The energy absorbed during a Jump of ‘e’ from lower level to higher level:
E=
1
𝜆
ℎ𝑐
𝜆
= Ef-Ei
= v = RyZ2
1
1
(𝑛2 − 𝑛2 )
𝑖
𝑓
Wave no.
When, Ry
=
Rydbergconstant
1.097 x 107 m-1
(3)
Absorption Spectral Lines for H-like atom.
n=∞
n=7
n=6
pfund
ni=5, nf = 6,7 -----------
n=5
n=4
Brackett
ni =4, nf =5,6----------n=3
Paschen
(ni=3, nf = 4,5,6------------------)
n=2
Balmer
(ni = 2,nf = 3,4,5,------------------)
n=1
Lyman
(ni = 1,nf = 2,3,4------------------)
We can calculate wave length of any particular spectral line. for example to calculate the wave length of 3 rd
member of paschen series .
ni =3,
nf = 4,5.6,7---------------1
1
1
= 𝑅𝛾 (9 - 36) =
𝜆
R =
12
𝑅𝛾
=
𝑅𝛾 3
36
=
𝑅𝛾
12
12𝑋10−7
1.097
= 1.093 X 10-6m
= 1.093 micro-meter
NUCLEII
Stability of a nucleus :
It is observed that middle size nucleii are most stable nucleii as compared to both
smaller and bigger nucleii in the periodic table. This is attributed to the fact that Binding energy per nucleon
(B.E/A) of both smaller and heavier nucleii is less as compared to that of middle size nucleii.
Thus, it is the binding energy per nucleon of differentiate nuclieii which determines the
stability larger the binding energy per nucleii larger will the stability. Hydrogen has the smallest value BE/A and is
therefore the most unstable element of periodic table, inspite of having only one proton inside the nucleus.
Binding Energy (BE): It is defined as the energy corresponding to the mass defect of different nucleii. It is
observed that the mass of a nucleus is always less than the sum of the masses of all the nucleons. This difference
in mass is called mass defect. According to Einstein mass energy equivalence (E=mc2) , the mass loss in the
formation of any nucleus can be assumed to be utilized in providing the binding force of the nucleus. This force
being one of the strongest force in nature is called nuclear force.
mass defect:
∆𝑚 = [{𝑚𝑝 z + (A-Z) 𝑚𝑛 } - MN]
mp
mass of the proton
mn
mass of the neutron
A
Atomic mass number
Z
Atomic Number
MN
Existing mass of the nucleus.
B.E = ∆mc2
In amu (atomic mass unit) :
As 1amu =1.67X10-27Kg
Energy corresponding to 1 amu ≡ 1.67 x 10-27 x c2
1 amu ≡
1.67 × 10−27 ×9 × 10+16
1.6 × 10−13
Mev
≡ 931.5 Mev
Nuclear Force: The force among different nucleon inside the nucleus is called nuclear force. Properties of
nuclear force are
(1)
Strongest force in nature.
(2)
Short range force i.e. it acts only upto a separation of few Fermi (1 fermi = 10-15m)
(3)
In-dependent of electrical changes. (n
n, n
p, p
p).
(4)
Always attractive in nature.
Difference between chemical and nuclear reactions:
Chemical Reaction
Nuclear Reaction
 Energy is evolved or absorbed during
*
Energy is always evolved.
reactions.
 It occurs due to participation of outer*
It occurs due to nucleons.
most electrons of the orbit.
 Small amount of energy is evolved or
*
Energy evolved is huge in magnitude.
absorbed.
 It is affected by external causes. (eg.
*
It remains unaffected by external causes.
temp. & pressure)
 After the reaction compounds of same
*
Components of entirely different
compositions are formed.
elementary structure are formed.
 Binding Energy Curve:
BE/A
•
Fe (Sb)
o
C
2.
U(238)
•
F
He
N
Li
1
1H
A
It is a graph b/w BE per nucleon and the atomic mass number of different nuclei. Following important
inferences can be drawn from the graph.
(a)
BE per nucleon of middle size nuclei is maximum as compared to that of smaller and bigger nucleii.
(b)
The smaller and the bigger nucleii convert themselves into middle size nucleii by way of nuclear fusion
and nuclear fission reactions respectively.
(c)
There exists peaks (local maximas) corresponding to the nucleii 2He4 ,6C12 and 8O16. These peaks indicate
the extra stability of above nucleii in their neighborhood.
Denstiy of Nucleii: It has been proved experimently that radius of different nucleii is directly proportion to
cube root of their atomic mass number i.e. __
R ∝ (A)1/3
=
R=R0 (A)1/3
Where RO
=
1.1 x 10-15m is a constant.
Nuclear density (𝜌) can be written as:
𝜌 =
𝜌 =
3𝑚𝐴
4𝜋𝑅03 𝐴
𝑚𝐴
4
3
( 𝜋𝑅 3 )
=
, Where m= mass of a proton, A= Atomic mass no.
𝜌=
3𝑚
4𝜋𝑅𝑜 3
Independent of the nature of the nucleii.
RADOACTIVITY:
The phenomena of spontaneous emission of 𝛼, 𝛽 & 𝛾(Photons) particles by heavier
elements in the periodic table is called radioactivity. It occure to provide stability of unstable nucleii.
LAWS of Radioactive Disintegrations:
(1)
Radioactive disintegration is an spontaneous process and is independent of external factors i.e.
temp & pressure
(It is impossible to predict about a particular nucleii breaking in a particular time. It might break
immediately and it might not for years to come.
(2)
No nucleii can emit both the 𝛼& 𝛽 particles together (Law of conservation of energy fails if it
happens. However after every𝛼𝑜𝑟 𝛽emission, emission, of 𝛾- photon (s) is a must. (This occurs to
bring the nucleii back to the ground state)
(3)
After the emission of 𝛼-particle the atomic no. of the parental nucleus is reduced by ‘2’ and the
atomic mass is reduced by ‘4’. Similarly after the emission of 𝛽-particle atomic no of the parental
nucleus is either increased of decreased by one and atomic mass remains the same .
(4)
The rate of decay of any radioactive sample is directly proportional to the no. of nucleii present in
the sample at that time. i.e.
𝑑𝑁
𝑑𝑡
∝𝑁
𝑑𝑁
𝑑𝑡
= - λN
Where ‘λ’ is decay constant measured in sec-1.
Derivation of the formula:
(a)
N = No 𝑒 −𝜆𝑡
−𝜆𝑡
(b)
𝑑𝑁
T1/2 =
0.693
𝑁
(c)
𝜆
𝑁0
1 𝑛
=( )
2
(d)
A= Ao𝑒
(a)
N = No 𝒆−𝝀𝒕 : According to the law of radioactive disintegration:
, A=
𝑑𝑡
Activity of radioactive sample.
𝑑𝑁
𝑑𝑡
= -λN
𝑁 𝑑𝑁
∫𝑁
0
𝑁
𝑡
= - ∫0 𝜆 𝑑𝑡
(𝑙𝑛 𝑁)𝑁
𝑁0 = - λ (t-o)
𝑁
lnN-lnNo= - λ t
(b)
= 𝑒 −𝜆𝑡
𝑁0
𝑁
𝑁0
N= 20
At t = T1/2
2
N = No𝑒 −𝜆𝑡
= N0𝑒 −𝜆𝑡
1
ln(2) = ln𝑒 −𝜆𝑡1/2 (ln1 – ln2) = - λt1/2
0- 2.303 log 102 = - λt1/2
t1/2 =
(c)
0.3013 ×2.363
𝜆
= 0.693
𝜆
After one half life. the no. of nucleii remaining
N -----------------
𝑁0
2
1
= N0(2)
After two half life- the no. of nucleii remaining N=
𝑁0
4
1 2
= (2) N0. Similarly after ‘n’ half life no. of
nucleii remaining is
1 𝑛
N = N0(2) =
(d)
𝑑𝑁
𝑑𝑡
A= A0𝒆−𝝀𝒕 :
𝑁
𝑁0
1 𝑛
=(2)
As N = N0𝑒 −𝜆𝑡
= - λ No𝑒 −𝜆𝑡
𝑑𝑁
By defn 𝑑𝑡 = - λ N = A No
A = Ao𝑒 −𝜆𝑡
t
Ao
- λ N o = Ao
t
Average Life: It is defined as the total life time of all present in the sample divided by the total no. of
nuclei in the sample.i.e.
tav =
𝑁
∫0 0 𝑡 𝑑𝑁
As: At t = 0
𝑁0
As N = No𝑒 −𝜆𝑡 dt
dN = -λ N0𝑒 −𝜆𝑡 dt
At N = N0 t = 0 & At N=0 t= ∞
tav =
−𝜆N0
𝑁0
0
∫∞𝑡 𝑒 −𝜆𝑡 dt
∞
tav = 𝜆 ∫0 𝑡𝑒 −𝜆𝑡
Let the total no. of nucleii
present in a sample be No. At time ‘t’ . Let the no.
of nucleii be N. If ‘dN’ nucleii get decayed just
now, then the life of dN nucleii can be written
as ‘tdN’
It can be solved using integration by part.
1
Units of Activity: Becquerel (Bq). 1 Bq = 1 decay per sec.
curie (Ci)
1 Ci = 3.7 x 1018 Bq.
tav = 𝜆
Explanation of Emission of 𝜶- particles:
consider a nucleus ZXA, getting converted to another nucleus
A-4 after the emission of a 𝛼- particle.
Z-2Y
A
A-4 + Q,
Where Q is the energy released during the reaction.
ZX
Z-2 Y
4
-2He
According to LOCO𝐿M: (Law of conversion of Linear momentum)
O = 𝑚𝛼 𝑣𝛼 + 𝑚𝛾 𝑣𝛾
𝑚𝛼 𝑣𝛼 = - 𝑚𝛾 𝑣𝛾
- - - - - - - - - - - - -(1)
Applying LOCO𝐾E: ( Law of conservation of Kinetic energy)
1
1
Q = 2 𝑚𝛼 𝑣𝛼2 + 2 𝑚𝛾 𝑣𝛾2
- - - - - - - - - - - - - - - - - - - - - (2)
From (1) & (2)
1
1
𝑚𝛼 𝑣𝛼
Q = 2 𝑚𝛼 𝑣𝛼2 + 2 𝑚𝛾 (
1
Q = 2 𝑚𝛼 𝑣𝛼2 [1 +
K.𝐸𝛼
𝑚𝛾
] = K.𝐸𝛼 [
𝑚𝛾 𝑄
=
K.𝐸𝛼 =
𝑚𝛼
𝑚𝛾
2
)
𝑚𝛾 + 𝑚𝛼
K.𝐸𝛼 =
𝑚𝛾 + 𝑚𝛼
𝑚𝛾
]
(𝐴−4)𝑄
𝐴−4+4
(𝐴−4)𝑄
𝐴
Example of 𝜶 - emission :29U238 -𝛼 90Th234+Q
Explanation of Emission of 𝜷– Particles: There are two kinds of 𝛽 particles. The one called –ve 𝛽 particle
is emitted by a nucleus when a neutron in the nucleus is converted into proton eg.
1
1
0
On
1P + -1𝛽 +𝛾(Antineutrino)
During –ve 𝛽 particle emission, another energetic charge and massless particle is always emitted called
antineutrino Another type of 𝛽 emission is that of emission of positron. Itis emitted by a nucleus when a
proton in the nucleus in converted into neutron eg.
1 1
0 + 𝛾, Here one proton is converted into a neutron and a positron (-ve electron ) and
(4)
1P On ++1𝛽
along with this another energetic particle called neutrino.
The neutrino nyhypothesis was firstever proposed by Pauli He pointed out that energy of 𝛽particles can’t
be continuous in absence of energetic mass less and charge less Particleenergy of 𝛽particles being in
quantum does not support the continuous graph (see. the fig.) obtained exponentially b/w. No .of
𝛽particles and the energy emitted . The need of gap filler was strongly felt by Pauli in his famous
hypothesis and it was later on experimentally verified.
No. of 𝛽 Particles emitted
End Pole Energy
Energy
K. Capture: During the emission of –ve 𝛽 particle the nucleus is left in excited state. In fact this is a
charged nucleus (Known as nuclide) with an extra proton as compared to the no. electrons in the orbits.
This charged nucleus usually captures an electron from k shell and the energy is released in the form of xray. This loss of electron is compensated by the atom capturing an electron from the Lattice. This
phenomena is called k.Capture:
Example :27CO60
- 𝛽 28Ni60 + 𝛾 (Photons)
Explanation of 𝜸 – emission: During the emission of ‘𝛼’ & ‘𝛽’ particles, the nucleii are generally left in
excited state. The transition of these nucleii from excided state to ground state evolves the emission of 𝛾photons.
60Co
EXP.
27
-1𝛽
0
60
28Ni
𝛾1=1.17 Mev
60
28Ni
𝛾2= 1.33 Mev
60
28Ni
Dual Nature:light behaves like waves as well as particle. This nature of light is called dual nature. It has
been experimentally proved that even matter possesses dual nature. It was de- Broglie who initiated the
concept of dual nature whichwas experimentally verified by Davission &Germerin their famous
experiment.
Expression of wavelength of Electron beam (matter wave): Consider and electron beam accelerated
through a potential ‘V’. The kinetic energy of a electron can be written as
1
mv2 = eV
2
From de- Broglie Equation
2𝑒𝑉
𝑣=√ 𝑚
----------------------------------------------- (1)
ℎ
λ = 𝑚𝑣 =
ℎ
2𝑒𝑉
𝑚√
𝑚
=
ℎ
√2𝑒𝑚√𝑉
Putting h = 6.034 X10-34Js
m= 9.1X 10-31 Kg & e = 1.6 X 10-19J
we get
0
λ=
12.27 𝐴
√𝑉
Davisson Germer Experiment :- with the help of this experiment we verify the wave nature of matter
wave. The experimental setup consists of an electron gun connected with a low tension battery. The
electrons emitted by the electron gun with almost zero K.E. is further accelerated by a potential
difference ‘V’. The collimated electron beam suffers Braggs differentiate (If it behaves like wave as
diffraction is purely a phenomena supporting the wave nature)
Atomic layer
𝜃
F
∅
<T
V
Ni crystal
D=Detector
𝜃
D
F= Filament
∅ = Angle of Diffraction
𝜃 = Glancing angle
Davisson and Germer collected the data involving intensities of diffracted electron beams vs the angle of
diffraction at-different accelerating potentials applied to the anode. They plotted various polar graphs as
shown below:
∅ = 500
∅
∅
I
I
I
48V
54V
I
58V
64V
The maximum contrast in the intensity was observed at anode voltage of 54V and diffraction angle
∅
0
= 50 . It is possible to have a maximum in the intensity only if there is a diffraction, following the Braggs
law: . The Wavelength of Electron beam (using Braggs law):
2d Sin 𝜃 = λ
As d = 0.91A0{for Ni}
As 2 𝜃+ ∅ = 180
𝜃= (
2X0.91A0XSin 650 = λ
180−50
2
) = 650
λ = 1.65 A0
It we calculate the same wavelength using the formula: λ=
λ = 1.66 A0
12.27𝐴0
√𝑉
, As V = 540
It must be noticed that the two calculations are producing the same result although the two formulae
used in the calculations are derived using two different nature of electron beams. The Braggs diffraction
on one hand supports the wave nature and the expression λ=
12.27𝐴0
√𝑉
is obtained using particle nature
of electron beam. This establishes the wave particle duality of matter waves.
Photoelectric Effect:
The phenomena of emission of electrons from a metallic surface after falling of a light beam of suitable
frequency is called photo-electric effect.
Laws of photoelectric effect:
(1)
For a given metal the emission of photoelectrons is not possible if the frequency of falling photons
is less than a critical value called threshold frequency. The work done in just knocking out an
electron from a metallic surface is called work function
Wo = hvo , vo = Threshold frequency.
(2)
The no. of photoelectrons ejected-out persec is directly proportional to the intensity of the photon
beam and is independent of the frequency of the photons.
(3)
The maximum K.E of the emitted photoelectron is independent of the intensity of the photon
beam and is dependent upon the frequency of the beam.
(4)
The photoelectric emission is an instantaneous process and there is no time lag between falling of
photons on the metallic surface and the emission of photo electrons.
Einsteins photoelectric Equation: Applying the law of conservation of energy, if ‘v’ be the frequency of
falling photons and vo be the threshold frequency, we can write:
1
2
hv –hvo = K.Emax = 2 𝑚𝑣𝑚𝑎𝑥
+ve
hv
-ve
K.Emax = h(v-vo)
---------------------------------- (1)
e
If the anode is given –ve potential, the photoelectrons
e
reaching the anode have to do work to overcome the
Rh
MA
electrostatic force. The minimum anode potential
Vs
(-ve) which can stop even the maximum energetic
electrons to reach the anode is called stopping potential (Vs). Thus we can also write,
eVs = K.Emax = hv - hvo
Einsteins equation of photoelectric effect.
--------------------------------
(2)
Graphical Explanation of laws of photo electric effect:
From the Einsteins 𝐸𝑞𝑛 :
eVs = hv-hvo
Vs
ℎ
𝑤
Vs = 𝑣 - 𝑜
ℎ
𝑒
𝑒
Slope = 𝑒
−
𝑤𝑜
Vo
𝑒
v
Y= mx + c
The graph shows that—
(i)
The emission of photoelectrons is not possible. If the frequency of photons is less that threshold
value. (1st law)
(ii)
As the frequency increases the stopping potential also increases linearly. Thus the k.E of the
emitted photoelectrons depends upon the frequency. (III law)
Note: Einstein used the above graph to calculate the value of plank’s constant. He did it by measuring
the slope of the graph.
Since slope = h/e, knowing ‘e’ h can be calculated.
Other three graphs (Experimently obtained) are as shown below
Photoelectric current
I1
I2
𝑣
(i)
𝑣
Photoelectric current
(ii)
I1>I2
𝑣
Anode Potentialvs
Intensity
Photoelectric current
V3>V2>V1
(iii)
The above graphs also supports the Einsteins
laws of photoelectric effect graph (i) & (ii)
V3
shows that the K.E depends on the frequency
V2
and the photoelectric current depends on the
V1
Anode Potential
intensity. Graph (ii) also supports the same
V3 V2 V1
statement.
Note: During numerical calculation children are advised to use the following shortcuts to reduce the
calculation time.
(i)
If stopping potential is 5Volt-the maximum K.E of the electons emitted will be 5ev
(ii)
ℎ𝑐
𝜆𝑒
=
i.e.
1243
𝜆
(ev-nm)
If λ = 400nm
ℎ𝑐
ℎ𝑐
Vs = 𝜆𝑒 -𝜆
0𝑒
ℎ𝑐
𝜆𝑒
1243
= 400 ev = 3.1075 ev
1
1
= 1243 (𝜆 − 𝜆 ), λ & λ0
0
are to be taken in nm . to express the result
Applications of photoelectric effect :(i)
Photo emissive cell
(ii)
Photo- voltaic cell
(iii)
Photo- conductive cell
*
*
*
*
uses
In television camera
In counting devices
In Burglar Alarm
In automatic doors.