3. # Thomson Model of Atom (1898)
⑦ -Electron
⑦ ⑦ -
Positively Charged
⑦
⑦ ↓
Matter
⑦ ⑦ ⑦
⑦ ⑦
⑪ Plum
Pudding Model
⑪ An atom consists of a
positively charged sphere with 10
embedded in it .
⑪ An atom as a whole is
electrically neutral .
6. Reford's a rations
-
- Most of the x-particles pass through the
gold foil without
any
L
deflection .
This shows that most of the space in an atom is
empty.
* Few
trely charged al
a particles defected through all a
large angles :
This shows that there is presence of positive
-
centre in the atom :
This positive centre is known as Nucleus .
X- very few a particles (bin 800) suffers deflection of 180 . This
shows that size of nucleus is very small (dense) .
/
↳ The volume occupied by the nucleus is
negligible compared to
the total volume of the atom .
This shows that radius of atom is
much
higher than that of nucleus . (size of nucleus is
goo times
me
the size of atomo
#e8 revolve around the nucleus.
7. Limitation of Thomson's Atomic model
It could not
explain large angle scattering of -particles.
~ failed to
explain the
stability of an atom .
⑪ It could not explain the
origin of
Peseries of
hydrogen and other atoms,
8. Impact Parameter (D) Veveimy3 .
The perpendicular distance of the
velocity vector of icle from
- - -
the central line of the nucleus ,
when the particle is far
away
-
from the nucleus is called impact parameter .
b =
1
zeot(02)
ATTE,
K
2 atomic Number
K
=
kinetic
energy .
- me
OF
angle of scattering .
9. Distance of Closest Approach ((o)
At the distance of closest approach whole kinetic
energy of
alpha particle is converted into potential energy
:
% =
2
zee
O
2 = atomic Number
me
K
=
kinetic
energy.
~
-
les
a-parl
- -
③
-
5kr Nucleus
*
K.
E
=
P.
E
-ea
e
11. # limitations of Rutherford's Atomic Model
↳ Rutherford's atomic model failed to
explain the
eity
of electrons
in a circular path.
m
e
↳ Rutherford atomic model cannot explain atomic line spectrum.
n
e
e
12. ⑪ The Bohr
Hydrogen Atom (Bohr's Model of atom)
↳
Stationary Orbit
ener w
· An electron moves
only in Certain circular orbits
"In
stationary orbit electron does not emit radiation
~ There is definite energy associated with each stable orbit .
~ An atom radiates
energy
~ only when it makes a transition
- mee
from one of these orbits to another ·
- - -
<Energy is radiated in the form of a photony ·
S
-
exicted state
E= hy
round
13. - - -
- nee nee
-
-
- - -
-
-
-
- -
-
- - - - o
-
L= MY L = R b
mir= ⑪
-
-
2x
n
e
e
e - -
e - e e
&
15. Derivation
ACC to Bohr's (H-atom) <Ip,
1e]
⑧
my Fe ratee
F =
- -
-
+
Ve A I 4,2 p2
⑭
Fo
↳
- . . .
-- 2
=
r
E r
My
-
r
=
4mi
using
value of y
- 27
137302=
0 . 529 e
r =
-
em te
- STMe
r
=
0 . 529A
vertimee set
In =
0 .
329
Ax Radiusotoht
16. velocity of e0
=
r
x =
here
H =
2=
1
x =
E *=
1
y =
e2 x =
e
=
2:
18x10
-
2nhSo
Y
=
20
18x10xvadity
R
=
Rydberg's constant
R
=
10097 X107m'
17. Ey ToG =
K.
E + P.
2
x =
E
kinetic
energy-
mi Potential Energy
=>
= (g)
v =
E Un
=
20
imzez
W = -
ze
- -
k =
1mxzes 45TE
me*
- --
-ne
!)
u = -
ze inze=
-ne
I 4 E
-
en
H-(2=
1)
H-atom 2=
1
k
=
me4 Eng
-
Ench
18. 4 2
TE =
k .
G + P.
E k
=
me" O = -
M2 2
-
14
Ench trin'
E =
mec -
Mee
e
14= -
E E
=
-
me e
↳Sh Or
E =
meme
E = -
K
E =
U
8222 -E
E
=
-
Me
85h
I=1
Emer
=
E =
-
13 i
I n2
19. # spectral series of Hydrogen atom
X
↳ discrete
When the electron in a
Hom jumps from
Higher energy
- - -
level to
eenergy level , the difference of energies of
-
the two
energy levels is femed as
lation
of
cular
line .
wavelength ,
known as nee ↓
e8 e
Higher energy State
(Ei)
N
- E=
hf
Ef
Ef E =
-
Meeten
Lif=
E
, -Ef
ht =
- m" + in-
⑧Enh ↳Ent
by
=
me"[ii]
21. Energy Level Diagram E = -
13. 6 eX
T2
Lyman Brackett Pfund
n =
7 series series series
- 0 .
28 eV
n =
6 -
O 38 eV
n
=
5 *
0.ee
n =
4 * *
n =
3
* IF Y -
151eV
Paschen
series
n
=
2 vvvvv -
3.
40 eV
Balmer
series
ground state
n =
1 *** V -
13 .
6eV
22. LO ->
rq
=
1
R = 2
,
3,
4---
Bhai -
nf
=
2
4
=
3 ,
- - -
Pani-- 4
=
3 Hi
=
4 ,
- - -
Bi - -
n
=
4 4 =
5 ----
Piyo ->
ng=
5
n
=
6 . - - - -
23. ↑
=
R
-
]
⑪
Lyman series
n =
1 n = 2
,
3
,
4,
5 . . -
[U.
Y region]
⑪ Balmer series
visible region?
nf
=
2 4
,
=
3
,
4,
5-----
Low
~ Paschen Series
[ infrared
region] xmp
n
=
3 4
,
= 4,
5 -----
⑭ Brackett Series
(infrared region]!
Low
nf
=
4 1
,
=
3,
6
,
7-- ...
High
⑪ pfund Series <infrared region ?
nq
=
5 4
p
=
6
,
7- - - -
0
24. Yo .
of spectral line
n = n
=
1
N =
(4=-n , ) (H2-
n
,
+
1)
- N=Best
See
2
25. Ionization
The process of knocking an electron out of the atom
is called Ionization :
Ionization
energy
The
energy required ,
to knock an electron
completely out
of the atom ·
Ionization
energy
=
13 EV
n2
26. DE BROGLIE'S EXPLANATION OF BOHR'S SECOND
POSTULATE OF QUANTISATION
MY =
h
e-Broy e ->
kave like
25
properties
stationary Orbit
↓
No emission of radiation.
Orbit
r
55 ↳ No wave
or
e
4 No
energy
loss .
2strn
=
nx
2st=
my
MY =
↳
EST
