Theory of Atomic Structure for Neet 2017

A NAME IN CONCEPTS OF PHYSICS

9027187359, 7351266266 XI &XII (CBSE & ICSE BOARD)

IIT-JEE / NEET /AIIMS / JIPMER / uptU

VARIOUS MODELS FOR STRUCTURE OF ATOM Dalton's Theory Every material is composed of minute particles known as atom. Atom is indivisible i.e. it cannot be subdivided. It can neither be created nor be destroyed. All atoms of same element are identical physically as well as chemically, whereas atoms of different elements are different in properties. The atoms of different elements are made up of hydrogen atoms. (The radius of the heaviest atom is about 10 times that of hydrogen atom and its mass is about 250 times that of hydrogen). The atom is stable and electrically neutral. Thomson's Atom Model: Thomson gave the first idea regarding structure of atom. According to this model. (i) An atom is a solid sphere in which entire and positive charge and it's mass is uniformly distributed and in which negative charge (i.e. electron) are embedded like seeds in watermelon. The positive charge and the whole mass of the atom is uniformly distributed throughout the sphere. electron –

–

positively charged matter

–

–

–

Positively charged sphere Electron

–

Success and failure Explained successfully the phenomenon of thermionic emission, photoelectric emission and ionization. The model fail to explain the scattering of - particles and it cannot explain the origin of spectral lines observed in the spectrum of hydrogen and other atoms. Shortcomings of Thomson's model (i) The spectrum of atoms cannot be explained with the help of this model (ii) Scattering of –particles cannot be explained with the help of this model n =

(iii) Angular frequency of electron in nth orbit (n) :

8 2 K 2 Z 2 e 4 m  n3 h3

n 

Z2 m n3

RUTHERFORD ATOM MODEL Rutherford experiments on scattering of - particles by thin gold foil Rutherford performed experiments on the scattering of alpha particles by extremely thin gold foils and made the following observations vacuum

beam of -particle





most -pass through

some are deviated through large angle 

THEORY NOTES FOR IIT - PMT

b

r0 Nucleu s

source of -particle about 1 in 8000 is repelled back



4 N()  cosec 

N()

ZnS gold foil screen 10–7m

e op sc cro mi

lead box

lead screen

90°

180°

-particle (Energy E )



ATOMIC STRUCTURE

P.L. SHARMA ROAD, center

SHASTRI NAGAR center CENTRAL MARKET,

Opp. Sagar Complex Meerut

OPP. SUMIT NURSING HOME, 1ST FLOOR AIM INTERNATIONAL Page 1

9027187359, 7351266266


XI &XII (CBSE & ICSE BOARD)


(i) Most of the -particles pass through the foil straight away undeflected. (ii) Some of them are deflected through small angles. (iii) A few -particles (1 in 1000) are deflected through the angle more than 90 o. (iv) A few  -particles (very few) returned back i.e. deflected by 180o. (v) Distance of closest approach (Nuclear dimension) The minimum distance from the nucleus up to which the -particle approach, is called the distance of closest approach (r0). From figure r0 

1 40

.

Ze 2 1 ; E  mv 2  K.E. of -particle 2 E

After Rutherford's scattering of -particles experiment, following conclusions were made as regard as atomic structure : (a) Most of the mass and all of the charge of an atom concentrated in a very small region is called atomic nucleus. (b) Nucleus is positively charged and it's size is of the order of 10–15 m  1 Fermi. (c) In an atom there is maximum empty space and the electrons revolve around the nucleus in the same way as the planets revolve around the sun. According to Rutherford scattering formula, the number of - particle scattered at an angle by a target are given by

Atom +

Nucleus

10–15 m

10–10 m

N0 nt(2Ze 2 )2 1 Size of the nucleus = 1 Fermi = 10–15 m N =  2 2 2 2 4(40 ) r (mv 0 ) sin 4  Size of the atom 1 Å = 10–10 m 2 Where N0 = number of - particles that strike the unit area of the scatter n = number of target atom per m3 t = thickness of target Ze = charge on the target nucleus 2e = charge on - particle r = distance of the screen from target v0 = velocity of - particles at nearer distance of approach the size of a nucleus or the distance of nearer approach is given by.

(vi) Impact parameter (b) : The perpendicular distance of the velocity vector ( v ) of the -particle from the centre of the nucleus when it is far away from the nucleus is known as impact parameter. It is given as

b

Note

:

Ze 2 cot( / 2)  b  cot( / 2) 1 2 40  mv  2 

 If t is the thickness of the foil and N is the number of -particles scattered in a particular direction ( =

constant), it was observed that

N t N  constant  1  1 . N2 t2 t

 Number of scattered particles : N 

1 sin 4 ( / 2)

Draw backs (i) Stability of atom : It could not explain stability of atom because according to classical electrodynamics theory an accelerated charged particle should continuously radiate energy. Thus an electron moving in an circular path around the nucleus should also radiate energy and thus move into smaller and smaller orbits of gradually decreasing radius and it should ultimately fall into nucleus.

e–

(ii) According to this model the spectrum of atom must be continuous where as practically it is a line spectrum. (iii) It did not explain the distribution of electrons outside the nucleus.

Instability of atom


ATOMIC STRUCTURE








(3) Bohr's model Bohr proposed a model for hydrogen atom which is also applicable for some lighter atoms in which a single electron revolves around a stationary nucleus of positive charge Ze (called hydrogen like atom) Bohr's model is based on the following postulates. (i) The electron can revolve only in certain discrete non-radiating orbits, called stationary orbits, for which total angular momentum of the revolving electrons is an integral multiple of

h ( ) 2

 h    mvr; where n = 1, 2, 3, ……..= Principal quantum number  2 

i.e. L  n 

(ii) The radiation of energy occurs only when an electron jumps from one permitted orbit to another. When electron jumps from higher energy orbit (E1) to lower energy orbit (E2) then difference of energies of these orbits i.e. E1 – E2 emits in the form of photon. But if electron goes from E2 to E1 it absorbs the same amount of energy. E1

E1 E1 – E2 = h

E2

E1 – E2 = h

E2 Absorption

Emission

Note :  According to Bohr theory the momentum of an e 

revolving in second orbit of H 2 atom will be

h



 For an electron in the n orbit of hydrogen atom in Bohr model, circumference of orbit  n ; th

where  = de-Broglie wavelength.

Bohr's Orbits (For Hydrogen and H2-Like Atoms). (1) Radius of orbit For an electron around a stationary nucleus the electrostatics force of attraction provides the necessary centripetal force i.e.

(Ze)e mv 2  40 r 2 r 1

……. (i)

also mvr 

nh 2

…….(ii) r

th

From equation (i) and (ii) radius of n orbit

rn 

n2h2 0 n 2 h2 n2   0 . 53 Å Z 4 2kZme2 mZe 2

 1   where k   4 0  

 rn 

n2 Z

n

Note :  The radius of the innermost orbit (n = 1) hydrogen atom (z = 1) is called Bohr's radius a0 i.e. a 0  0.53 Å (2) Speed of electron : From the above relations, speed of electron in nth orbit can be calculated as

vn 

2

v

2

2kZe Ze  c  Z 6 Z   m / sec .  2.2  10 nh 2 0 nh  137  n n

where (c = speed of light 3  108 m/s)

Note :  air is equal to

n

The ratio of speed of an electron in ground state in Bohr's first orbit of hydrogen atom to velocity of light in

e2 1 (where c = speed of light in air)  2 0 ch 137


ATOMIC STRUCTURE








(3) Some other quantities For the revolution of electron in nth orbit, some other quantities are given in the following table Quantity (1) Angular speed

(2) Frequency

(3) Time period

n 

vn mz 2 e 4  rn 2 02 n 3 h 3

n 

n 

n mz 2 e 4  2 4 02 n 3 h 3

n 

Tn 

(4) Angular momentum

(5) Corresponding current

(6) Magnetic moment

1

n



4 02 n 3 h 3 mz 2 e 4

 h  Ln  mvn rn  n    2 

i n  e n 

mz 2 e 5 4 02 n 3 h 3

 

M n  in A  in  rn2 (where  0 

(7) Magnetic field

Dependency on n and Z

Formula

B

 0 in 2rn



Tn 

Z2 n3 Z2 n3 n3 Z2

Ln  n

in 

Z2 n3

Mn  n

eh  Bohr magneton) 4m

m 2 z 3 e 7  0 8 03 n 5 h 5

B

Z3 n5

(4) Energy (i) Potential energy : An electron possesses some potential energy because it is found in the field of nucleus potential energy of electron in nth orbit of radius rn is given by U  k.

(Ze)(e) kZe 2  rn rn

(ii) Kinetic energy : Electron posses kinetic energy because of it's motion. Closer orbits have greater kinetic energy than outer ones. As we know

mv 2 k. (Ze)(e) kZe 2 | U |   Kinetic energy K   rn 2rn 2 rn2

(iii) Total energy : Total energy (E) is the sum of potential energy and kinetic energy i.e. E = K + U  E

n 2 h 2 0 kZe 2 also rn  . 2rn mze 2  me 4  8 2 h 2  0

Hence E    where R 

 z2  me 4 .     n2  8 2 ch3   0

 z2 Z2 Z2  ch   R ch   13 . 6 eV  n2 n2 n2 

me 4 = Rydberg's constant = 1.09  107 per metre 8 02 ch3

Note : 

Each Bohr orbit has a definite energy

 For hydrogen atom (Z = 1)  En  

13.6 eV n2

 The state with n = 1 has the lowest (most negative) energy. For hydrogen atom it is E1 = – 13.6 eV.

– 2.17  10 18 J ~ – 31.6 eV .  Rch = Rydberg's energy ~  E  K 

U . 2


ATOMIC STRUCTURE






A NAME IN CONCEPTS OF PHYSICS IIT-JEE / NEET /AIIMS / JIPMER / uptU

(iv) Ionisation energy and potential : The energy required to ionise an atom is called ionisation energy. It is the energy required to make the electron jump from the present orbit to the infinite orbit.



Hence Eionisation  E  En  0    13.6 



13.6 Z 2 Z2    eV n2 n 2 

For H2-atom in the ground state Eionisation 

 13.6(1) 2  13.6 eV n2

The potential through which an electron need to be accelerated so that it acquires energy equal to the ionisation energy is

Vionisation 

called ionisation potential.

Eionisation e

(v) Excitation energy and potential : When the electron is given energy from external source, it jumps to higher energy level. This phenomenon is called excitation. The minimum energy required to excite an atom is called excitation energy of the particular excited state and corresponding potential is called exciting potential.

EExcitation  EFinal  EInitial and VExcitation 

Eexcitation e

(vi) Binding energy (B.E.) : Binding energy of a system is defined as the energy released when it's constituents are brought from infinity to form the system. It may also be defined as the energy needed to separate it's constituents to large distances. If an electron and a proton are initially at rest and brought from large distances to form a hydrogen atom, 13.6 eV energy will be released. The binding energy of a hydrogen atom is therefore 13.6 eV.

Note :  For hydrogen atom principle quantum number n 

13.6 . (B.E.)

(5) Energy level diagram The diagrammatic description of the energy of the electron in different orbits around the nucleus is called energy level diagram.

Energy level diagram of hydrogen/hydrogen like atom

Note : 

n=

Infinite

Infinite

E = 0 eV

n=4

Fourth

Third

E4 = – 0.85 eV

– 0.85 Z 2

+ 0.85 eV

2

+ 1.51 eV

0 eV

n=3

Third

Second

E3 = – 1.51 eV

– 1.51 Z

n=2

Second

First

E2 = – 3.4 eV

– 3.4 Z 2

n=1

First

Principle quantum number

Orbit

Ground Excited state

0 eV

+ 3.4 eV 2

E1 = – 13.6 eV

– 13.6 Z

Energy for H2 – atom

Energy for H2 – like atom

+ 13.6 eV Ionisation energy from this level (for H2 – atom)

In hydrogen atom excitation energy to excite electron from ground state to first excited state will be

 3.4  (13.6)  10.2 eV . and from ground state to second excited state it is [  1.51  (13.6)  12.09 eV ].  In an H 2 atom when e  makes a transition from an excited state to the ground state it’s kinetic energy increases while potential and total energy decreases.


ATOMIC STRUCTURE








(6) Transition of electron When an electron makes transition from higher energy level having energy E2(n2) to a lower energy level having energy E1 (n1) then a photon of frequency  is emitted

    13.6 Z 2  1  1   n2 n2 2   1  1 E E 2  E1 1  (ii) Frequency of emitted radiation : E  h      Rc Z 2  2  2  h h  n1 n2  E2  Rc h Z 2

(i) Energy of emitted radiation E  E 2  E1 

n22

 Rch Z 2    n12 

    n2

(iii) Wave number/wavelength Wave number is the number of waves in unit length

 

1





 c



 1 1  13.6 Z  RZ 2  2  2    hc  n1 n2  1

2

E,,

 1 1    n2  n2  2   1

E1

n1 Emission

(iv) Number of spectral lines : If an electron jumps from higher energy orbit to lower energy orbit it emits raidations with various spectral lines.

(n2  n1  1)(n2  n1 ) 2 n (n  1) th If electron falls from n orbit to ground state (i.e. n2 = n and n1 = 1) then number of spectral lines emitted N E  2 If electron falls from orbit n2 to n1 then the number of spectral lines emitted is given by N E 

Note :  Absorption spectrum is obtained only for the transition from lowest energy level to higher energy levels. Hence

the number of absorption spectral lines will be (n – 1). (v) Recoiling of an atom : Due to the transition of electron, photon is emitted and the atom is recoiled Recoil momentum of atom = momentum of photon  Also recoil energy of atom 

p2 h2  2m 2m 2

 1 1  hRZ 2  2  2   n1 n2 h

   

(where m = mass of recoil atom)

(7) Drawbacks of Bohr's atomic model (i) It is valid only for one electron atoms, e.g. : H, He+, Li+2, Na+1 etc. (ii) Orbits were taken as circular but according to Sommerfield these are elliptical. (iii) Intensity of spectral lines could not be explained. (iv) Nucleus was taken as stationary but it also rotates on its own axis. (v) It could not be explained the minute structure in spectrum line. (vi) This does not explain the Zeeman effect (splitting up of spectral lines in magnetic field) and Stark effect (splitting up in electric field) (vii) This does not explain the doublets in the spectrum of some of the atoms like sodium (5890Å & 5896Å)

Hydrogen Spectrum and Spectral Series: When hydrogen atom is excited, it returns to its normal unexcited (or ground state) state by emitting the energy it had absorbed earlier. This energy is given out by the atom in the form of radiations of different wavelengths as the electron jumps down from a higher to a lower orbit. Transition from different orbits cause different wavelengths, these constitute spectral series which are characteristic of the atom emitting them. When observed through a spectroscope, these radiations are imaged as sharp and straight vertical lines of a single colour. Photon of wavelength  +

+

Spectrum

+

Emission spectra


ATOMIC STRUCTURE








Spectral series The spectral lines arising from the transition of electron forms a spectra series. (i) Mainly there are five series and each series is named after it's discover as Lymen series, Balmer series, Paschen series, Bracket series and Pfund series. (ii) According to the Bohr's theory the wavelength of the radiations emitted from hydrogen atom is given by

1 1  R 2  2    n1 n2  1

where n2 = outer orbit (electron jumps from this orbit), n1 = inner orbit (electron falls in this orbit) (iii) First line of the series is called first member, for this line wavelength is maximum ( max) (iv) Last line of the series (n2 = ) is called series limit, for this line wavelength is minimum (min) continuum 5th excited state 4th excited state

n=6 n=5

3rd excited state

n=4

–0.38 eV –.54 eV

Pfund Series Brackett Series

–0.85 eV

Paschen Series –1.51 eV

2nd excited state n = 3 Balmer Series st

1 excited state

–3.40 eV

n=2

Lyman Series ground state

Spectral series

–13.6 eV

n=1

Transition

Wavelength () 

Maximum

n1

n12 n22 (n22

 n12 )R

wavelength

 n and n2  n  1

 max 

n 2 (n  1) 2 (2n  1)R

 max 

(1) 2 (1  1) 2 4  (2  1  1)R 3 R

1. Lymen series

n2 = 2, 3, 4 … n1 = 1

2.Balmer series

n2 = 3, 4, 5 … n1 = 2

n1 = n = 2, n2 = 2 + 1 = 3 36 max  5R

3. Paschen series

n2 = 4, 5, 6 … n1 = 3

n1 = n = 3, n2 = 3 + 1 = 4 144 max  7R

4. Bracket series

n2 = 5, 6, 7 …  n1 = 4

n1 = n = 4, n2 = 4 + 1 = 5 400 max  9R

5. Pfund series

n2 = 6, 7, 8 …  n1 = 5

n1 =  = 5, n2 = 5 + 1 = 6 900 max  11R




n12 2   1  n1  R  n22  

Region

 max (n  1) 2   min (2n  1)

Minimum wavelength

n2

 , n1  n

min 

n2 R

n1 = n = 1 1 min  R

4 3

Ultraviolet region

9 5

Visible region

n1 = n = 3 9 min  R

16 7

Infrared region

n1 = n = 4 16 min  R

25 9

Infrared region

36 11

Infrared region

 min 

min 

4 R

25 R

ATOMIC STRUCTURE






A NAME IN CONCEPTS OF PHYSICS IIT-JEE / NEET /AIIMS / JIPMER / uptU

Concepts



With the increase in principal quantum number the energy difference between the two successive energy level decreases, while wavelength of spectral line increases. E'  E' '  E' ' '

'  ' '  ' ' '

E, 

E  E'  E' '  E' ' ' 1 1 1 1     ' ' ' ' ' '



n=4

E, 

n=3

E, 

n=2 E,  n=1

Rydberg constant is different for different elements R( =1.09  107 m–1) is the value of Rydberg constant when the nucleus is considered to be infinitely massive as compared to the revolving electron. In other words, the nucleus is considered to be stationary.

R

In case, the nucleus is not infinitely massive or stationary, then the value of Rydberg constant is given as R' 

1

m

where m is

M

the mass of electron and M is the mass of nucleus.



Atomic spectrum is a line spectrum Each atom has it's own characteristic allowed orbits depending upon the electronic configuration. Therefore photons emitted during transition of electrons from one allowed orbit to inner allowed orbit are of some definite energy only. They do not have a continuous graduation of energy. Therefore the spectrum of the emitted light has only some definite lines and therefore atomic spectrum is line spectrum.



Just as dots of light of only three colours combine to form almost every conceivable colour on T.V. screen, only about 100 distinct kinds of atoms combine to form all the materials in the universe.

Ex.

A hydrogen atom in the ground state is excited by radiations of wavelength 975 Å. Find : (a) the energy state to which the atom is excited. (b) how many lines will be possible in emission spectrum

Sol.

(a)

 = 975 Å = 975 x 10-10 m

1 1 1 =R 2  2  1 n  (b)

1 1 1 = 1.1 × 107  2  2  10 975 10 1 n 



or

n=4

n=4 Number of spectral lines (N) = N=

n(n  1) 2

4  (4  1) =6 2

Ex.

Possible transition 4 3, 4  2, 4  1, 3  2, 3  1, 2  1 Find the longest and shortest wavelength when a hydrogen atom in the ground state is excited by radiations of wavelength 975 Å.

Sol.

=

hc eE



12400 Å E(eV)

for longest wavelength

max =

for smallest wavelength

min =

12400 12400 = 18787.8 Å  E 4 3 0.66 hc 12400 12400  973 Å   eE E 4 1 12.75

Ex.

How many lines will be possible in the absorption spectrum when a hydrogen atom in the ground state is excited by radiations of wavelength 975 Å.

Sol.

Here n = 4

Number of absorption lines be 1 2, 1  3, 1 – 4


ATOMIC STRUCTURE








Ex.

Find the first and second excitation potentials of an atom when its ionisation potential is 122.4 V.

Sol.

I.P. = 122.4 V

Ex.

122.4 = 108.8 V 9 Find the atomic number of atom when given that its ionisation potential is equal to 122.4 V.

Sol.

I.P. = 122.4 V

Ex.

If the second excitation potential of an atom is 108.8 V then, find (a) The recoil momentum of atom in kg-m/sec (b) The energy of recoil atom in joule.

Sol.

(a)

Eex2 = 108.8V

(b)

E

Eex1 = 122.4 –

122.4 = 91.8 V 4

Eex2 = 122.4 –

Ex.

E = Z2EH

Z=

p=

E 122.4 = =3 E H1 13.6

E 108.8 1.6 10 19 = = 5.8 × 10-26 kg-m/s c 3  10 8

5.82 10 52 p2  = 1.44 x 10-25 J 2M 2  7 1.67 10 27

Find the maximum wavelength of Brakett series of hydrogen atom.

Sol.

n1 = 4 and n2 = 5



1  max

1 1 =R 2  2 4 5 

or

max =

25  16  1010 = 40400 Å 9  1.1  107

Ex.

Find the ratio of wavelength of first line of Lyman series of doubly ionised lithium atom to that of the first line of Lyman series of deuterium (1H2).

Sol.

For deuterium (1H2)

l 1 1 = R × 12  2  2  × D 1 2 

Ex.

 Li l 1 1 1 = R × 32  2  2  = =1:9 9  Li D 1 2  Find the value of magnetic induction at the proton due to electron motion, if the radius of the first orbit of hydrogen atom is 0.5 Å and the speed of electron in it is 2.2 x 106 m/sec. r = 0.5 Å and V = 2.2 × 106 m/sec  ev 10 7  1.6  10 19  2.2  10 6 B= 0 2  = 14.08 Tesla 4 r 25 10 22 Find the ratio of equivalents current due to electron motion in first and second orbits of hydrogen atom.

Sol.

In 

For lithium (Li+2) Ex. Sol.

3

Ex.

I1  n2  2      =8:1 I 2  n1  1  Find the ratio of the area of orbit of first excited state of electron to the area of orbit of ground level for hydrogen atom.

Sol.

A  r2  n4

Ex.

1 n3



3

4

A2  2  16    = 16 : 1 A1  1  1 If the ionisation potential in the ground state for hydrogen is 13.6 e.V., then find the excitation potential of third orbit.



13.6  13.6  –  2  = 0.66 Ev 42  3  Ex. When an electron jump from fourth orbit to ground state of hydrogen atom then calculate the wavelength of emitted photon. 13.6 E = E2 – E1= – – (– 13.6) = – 3.4 + 13.6 = 10.2 eV 4 1240 (eV) (nm) hc The wavelength  = = = 121.6 nm.  122 nm. E 10.2 (eV)

Sol.

I.P. = 13.6 eV

E4 - E3 =


ATOMIC STRUCTURE





Theory of Atomic Structure for Neet 2017

Recommend Documents