A t o m i c S t r u c t u r e
CHEMISTRY
ATOMIC STRUCTURE
1. LAWS OF CHEMICAL COMBINATIONSLaw of conservation of mass: Laws of conservation of mass states that in a chemical reaction the weight of
products is always equal to the weight of reactants.
Law of definite proportions: Law of definite proportions states that the elemental composition of a compound
always remains same if it is analysed form various sources e.g. water from a river or ditch or pond either in India or in USA would always give H : O ratio as 2 : 1.
Law of Multiple Proportions: Law of multiple proportion states that elements combine in simple whole number
ratios to form various types of compounds e.g. The ratio of N : O is 1: 1, 1 : 2 and 2 : 1 in NO, NO2 and N2O. respectively.
2. DALTON’S THEORY OF ATOM
John Dalton developed his famous theory of atom is 1803. The main postulates of his theory were :
Atom was considered as a hard, dense and smallest indivisible particle of matter.
Each element consists of a particular kind of atoms.
The properties of elements differ because of differences in the kinds of atoms contained in them. This theory provides a satisfactory basis for the law of chemical combination.
Atom is indestructible i.e. it cannot be destroyed or created.
Drawbacks
It fails to explain why atoms of different kinds should differ in mass and valency etc.
The discovery of isotopes and isobars showed that atoms of same elements may have different atomic masses (isotopes) and atoms of different kinds may have same atomic masses (isobars).
The discovery of various sub-atomic particles like X-rays, electrons, protons etc. during the 19th century lead to the idea that the atom was no longer an indivisible and smallest particle of the matter..
3. DISCOVERY OF ELECTRON
William Crookes found that certain rays come out of cathode and gain lots of energy because of high acceleration potential before colliding with a gas molecule. This happens especially when the gas pressure is low. These rays are known as cathode rays.
Cathode
High Voltage
+
Gas at very low pressure
Cathode Anode Suction pump
Fig. Cathode ray Tube experiment
Detailed study of cathode rays by J.J. Thomson led to the discovery of electrons. He observed that
1.Cathode rays always travel in straight lines
2.They are negatively charged. Cathode rays turn towards positively charged plates. 3.Charge on particles constituting rays was determined by Oil
Drop experiment by Millikan as 1.6 10 19 coulomb.
4.Specific charge (e/m) does not change when the gas inside discharge tube was changed indicating that electron is fundamental particle.
e/m for electron = 1.758 10 11Coulombs kg/ .
5.Value of charge on electron = 1.6 1019 Coulombs
6.Mass of the electron = 1.9 10 31 kg
.
4. ANODE RAYS (DISCOVERY OF PROTON)
It is well-known fact that the atom is electrically neutral. The presence of negatively-charged electrons in the atom amphasized the presence of positively-charged particles.
To detect the presence of positively-charged particles, the discharged tube experiment was carried out, in which a perforated cathode was used. Gas at low pressure was kept inside the tube. On passing high voltage between the electrodes it was observed that some rays were emitted from the side of the anode. These rays passed through
the holes in the cathode and produced green fluorescence on the opposite glass was coated with ZnS. These rays consist of positively-charged particles known as protons.
+ ZnS Coating H2 gas inside at low pressure Perforated cathode
For anode rays, e/m is not fundamental property as different gases used have different mass on C-12 scale. Highest e/m is for hydrogen gas.
5. DISCOVERY OF NEUTRON
After the discovery of electrons and protons, Rutherford (1920) had predicted the existence of a neutral fundamental particle. In 1930, Bethe and Becker reported from Germany that if certain light elements, like beryllium, were expoesd to alpha rays from the naturally radioactive polonium, a very highly penetrating radiation was obtained. Similar results were obtained by Irene Cureia dn F. Jolit (1932). Chadwick (1932) demonstrated that this mysterious radiation was a stream of fast moving particles of about the same mass as a proton but having no electric charge. Because of their electrical neutrality, these particles were called neutron.
The lack of charge on the neutron is responsible for its great penetraing power.
Thus, a neutron is a sub-atomic fundamental particle which has a mass 1.675 10 24 g (approximately 1 amu),
almost equal to that of a proton or a hydrogen atom but carrying no electric charge. The e m/ value of a neutron is thus zero.
6. CONSTITUENTS OF ATOM
Some of the well known fundamental particles present in an atom are-protons, electrons and neutrons. Many others were discovered later viz positron, neutrinos etc.
Subatomic Symbol Unit Charge Unit Mass Charge in Mass in AMU
particles Coulomb Proton 1p1 +1 1 19 1.60 10 1.007825 Neutron 0n1 0 1 0 1.008665 Electron –1e0 –1 Negligible 19 1.602 10 4 5.489 10 Do you Know?
Chandwick in 1932 discovered the neutron by bombarding elements like beryllium with fast moving -particles. He observed that some new particles were emitted which carried no charged and had mass equal to that of proton.
7. THEORIES OF ATOM
After the discovery of electrons, protons and neutrons several theories were proposed by various scientists such as
1.Plum Pudding Model (Thomson Model of Atom) 2.Rutherford’s Model of Atom
3.Planck’s quantum theory 4.Bohr’s Atomic Model
Thomson Model of Atom
According to this model, electrons are embedded in uniform sphere of positive charge to confer electrical neutrality. This model was satisfactory to the extent that the electrostatic forces of repulsion among the electron cloud is balanced by the attractive forces between the positively charged mass and the electron. How ever, this model fails to explain the results of ionisation and scattering experiment and is, therefore discarded.
most of them are passed through the metal foil with little or no deviation. However a small proportion of -particles are scattered through large angles and even bounced back i.e. deflected through 1800. From these
observations, Rutherford drew the following conclusions:
Gold foil Zinc sulphide screen Reflected -particles
Lead plate Lead block
Fig. Rutherford's Experiment
(a) As most of the -particles passed through the foil without undergoing any deflection, there must be sufficient empty space within the atom.
(b) As -particles are positively charged, deflected by large angle there must be heavy small positively charged body present in the atom, which is called Nucleus.
From these observations, Rutherford proposed the following model of atom:
(a) Atom is composed of a positively charged nucleus where the whole mass of the atom is concentrated and the electrons are present in relatively large volume around the nucleus. The total positive charge carried by nucleus must be equal to the total negative charge carried by electrons and electroneutrality of atom is maintained.
Fig: Thomson Model electron positive sphere
(b) Electron are constantly moving around the nucleus in different orbits.
The centrifugal force arising from this motion balances the electrostatic attraction between the nucleus and the electron. Therefore the electron don’t fall into the nucleus.
Drawbacks in Rutherford Model
(a) The most fundamental objection arises from the electromagnetic theory of radiation which predicts that when a charged body moves in a circular path, it should radiate energy continuously. As the electron is a negatively charged particle revolving around the nucleus, it should radiate energy continuously. As a result, the electron should fall into the nucleus.
(b) As the electron is continuously radiating energy, the spectra should be continuous. Actually, the spectra of atom is a line spectra.
8. ATOMIC NUMBER OF AN ELEMENT
= Total number of protons present in the nucleus = Total number of electrons present in the atom
* Atomic number is also known as proton number because the charge on the nucleus depends upon the number of protons.
* Since the electrons have negligible mass, the entire mass o the atom is mainly due to protons and neutrons only. Since these particles are present in the nucleus, therefore they are collectively called nucleons.
* As each of these particles has one unit mass on the atomic mass scale, therefore the sum of the number of protons and neutrons will be nearly equal to the mass of the atom.
Mass number of an element = No. of protons + No. of neutrons.
* The mass number of an element is nearly equal to the atomic mass of that element. However, the main difference between the two is that mass number is always a whole number whereas atomic mass is usually not a whole number.
* The atomic number (Z) and mass number (A) of an element ‘X’ are usually represented alongwith the symbol of the element as
A Z Mass Number Atomic Number Symbol of the element
x
e.g. 23 35 11Na,17Cl and so on. 1.Isotopes:Such atoms of the same element having same atomic number but different mass numbers are called isotopes.
1 2 3
1H,1H and H1 and named as protium, deuterium (D) and tritium (T) respectively. Ordinary hydrogen is
protium.
2. Isobars:
Such atoms of different elements which have same mass numbers (and of course different atomic numbers) are called isobars.
e.g. 40 40 40
18Ar,19K,20Ca. 3. Isotones:
Such atoms of different elements which contain the same number of neutrons are called isotones.
e.g. 14 15 16
6C, 7K, 8O. 4. Isoelectronics:
The species (atoms or ions) containing the same number of electrons are called isoelectronic. For Example, O2,F,Na,Mg2, Al3,Ne
all contain 10 electrons each and hence they are isoelectronic.
Illustration.
Complete the following table:
Particle Mass No. Atomic No. Protons Neutrons Electrons
Nitrogen atom – – – 7 7
Calcium ion – 20 – 20 –
Oxygen atom 16 8 – – –
Bromide ion – – – 45 36
Solution.
For nitrogen atom.
No. of electron = 7 (given)
No. of neutrons = 7 (given)
No. of protons = Z = 7 (atom is electrically neutral)
Atomic number = Z = 7
Mass No. (A) = No. of protons + No. of neutrons = 7 + 7 = 14
For calcium ion.
No. of neutrons = 20 (given)
Atomic No. (Z) = 20 (given)
No. of protons = Z = 20;
No. of electrons in calcium atom = Z = 20
But in the formation of calcium ion, two electrons are lost from the extranuclear part according to the equation
2
2
Ca Ca e
but the composition of the nucleus remains unchanged.
No. of electrons in calcium ion = 20 – 2 = 18
Mass number (A) = No. of protons + No. of neutrons = 20 + 20 = 40.
For oxygen atom.
Mass number (A) = No. of protons + No. of neutrons = 16 (Given)
No. of electrons = Z = 8
No. of neutrons = A – Z = 16 – 8 = 8
Some important characteristics of a wave:
Crest Crest
Trough Trough
a a
Wavelength of a wave is defined as the distance between any two consecutive crests or troughs. It is represented
by and is expressed in Å or m or cm or nm (nanometer) or pm (picometer).
8 10
1 Å10 cm10 m
9 12
1nm10 m, 1pm10 m
Frequency of a wave is defined as the number of waves passing through a point in one second. It is represented
by v (nu) and is expressed in Hertz (Hz) or cycles/sec or simply sec–1 or s–1.
1 Hz = 1 cycle/sec
Velocity of a wave is defined as the linear distance travelled by the wave in one second. It is represented by v and
is expressed in cm/sec or m/sec (ms–1).
Amplitude of a wave is the height of the crest or the depth of the trough. It is represented by ‘a’ and is expressed
in the units of length.
Wave number is defined as the number of waves present in 1 cm length. Evidently, it will be equal to the
reciprocal of the wavelength. It is represented by v (read as nu bar). 1
v
If is expressed in cm, v will have the units cm–1.
Relationship between velocity, wavelength and frequency of a wave. As frequency is the number of waves passing through a point per second and is the length of each wave, hence their product will give the velocity of the wave. Thus,
v v
Cosmic rays < - rays < X-rays < Ultraviolet rays < Visible < Infrared < Micro waves < Radio waves.
9. PLANCK’S QUANTUM THEORY (1901)
It states
1.Radiant energy is emitted or absorbed discontinuously in the form of tiny bundles of energy called Quanta. 2.Each quantum is associated with a definite amount of energy E which is proportional to frequency of radiation.
Ehv
Where, h = Planck’s constant = 6.626 10 34 Joulesec
. v= Frequency of the light radiation
3.A body can emit or absorb radiations only in whole multiples of quantum i.e. Enhv where n 1, 2, 3...
v c
where c = velocity of light
E hc
9(a). Some Important Formulae
1.A = Z + N (Number of neutrons)
2.dynamic mass of particle 2 1/ 2
0/[1 ( / ) ] mm v c 3.Radius of nucleus 1/ 3 15 0( ) , 0 1.2 10 RR A R m 4. cv 5.wave number = v1/ 6.Ehvhc/hcv 7. 122 9 2 2 0 1 ; 9.0 10 / 4 q q F K K Nm C r 8. E hvE2E1 Illustration 1.
Calculate number of photon coming out per sec. from the bulb of 100 watt. If it is 50% efficient and wavelength coming out is 600 nm.
Solution.
energy = 100 J
energy of one photon =
34 8 19 9 6.625 10 3 10 6.625 10 2 600 10 hc no. of photon = 100 1019 15.09 1019 6.625 Illustration 2.
Certain sun glasses having small of AgCl incorporated in the lenses, on exposure to light of appropriate wavelength turns to gray colour to reduce the glare following the reactions:
( )
hv
AgClAg Gray Cl
If the heat of reaction for the decomposition of AgCl is 248 kJ mol–1, what maximum wavelength is needed to
induce the desired process?
Solution.
Energy needed to change = 248 10 3 J mol/
If photon is used for this purpose, then according to Einstein law one molecule absorbs one photon. Therefore,
A 248 103 hc N 34 8 23 7 3 6.625 10 3.0 10 6.023 10 4.83 10 248 10 m .
10. BOHR’S ATOMIC MODEL
The postulates of Bohr’s atomic theory regarding stability of electrons of an atom are as follows:
(i) The electrons is an atom revolve around the nucleus only in certain selected circular orbits. These orbits are known as energy levels or stationary states. An electron can be excited from a lower state to higher state with the absorption of a quantum of energy, or can come down from a higher to lower state with emission of a radiation of energy (as shown in figure) equal to energy to quantum E E2E1hv. E and 2 E are energies of the1
+
Absorption of E Emission of
radiation with energy E(E2 E1)
E1
Fig. Bohr's Atomic Model
(ii) The stability of the circular motion of an electron requires that the electrostatic force (due to the attraction between the nucleus and the electron) provides the necessary centripetal force for the motion of electron.
2 2 0 1 ( ) . / 4 Ze e mv r r ... (i)
where Z – atomic number
e – charge on electron
0
– permittivity of the charge in vacuum
r – distance between positive charge & electron
(iii) Angular momentum of electron is quantised i.e. electron can revolve only in those orbits where its angular momentum is an integral multiple of h/ 2
/ 2
mvrnh ... (ii)
where, v – velocity of electron m – mass of electron h – Planck’s constant
1, 2, 3...
n are known as Principal quantum number..
Bohr’s Atomic Radius
from (i) &(ii) we have
-from (i) 2 2
0
/(4 )
v ze mr ... (iii)
& from (ii) 2 2 2 2 2 2
/ 4
v n h m r ... (iv)
Equating (iii) and (iv), we have
2 2 2 2 2 2 0 /(4 ) / 4 Ze mrn h m r or, 2 2 0 2 2 (4 ) 4 n h r mZe
r is called Bohr’s radius.
Where h,, ,e m and are constants, Thus, 0
2
( / )
rK n Z
Putting the values of h,, ,e m and , 0
11 2
(5.297 10 ) ( / )
r m n Z
Illustration 1.
For hydrogen atom Z , therefore radius of the first orbit = 1 (5.297 10 11m) (1 /1)2 For He ion
2
Z , therefore radius of the first orbit = (5.297 10 11m) (1 / 2)2 11
2.649 10 m
Solution.
Substituting value of r in equation no. (ii) 2 0 2 ( / ) 4 Ze v K Z n nh
The energy of an electron in an orbit
E = Kinetic energy + Potential energy =
2 2 0 1 2 (4 )r Ze mv Putting (mv)2
from equation (i), we have
2 2 2 0 0 0 2(4 ) (4 ) 2(4 ) Ze Ze Ze E r r r
Now, putting the value of ' 'r , we have
2 2 4 2 2 18 2 2 2 2 2 0 2 ( / ) (2.18 10 ) ( / ) (4 ) mZ e E K Z n J Z n n h
The energy difference between two energy levels n2 and n1 is given by
2 1 2 2 4 2 2 1 2 2 2 0 1 2 1/ 1/ (4 ) n n mZ e E E E n n h 18 2 2 2 1 2 1 1 (2.18 10 J Z) n n
It terms of wave-number, we have
2 2 2 1 2 1/ 1/ H v R Z n n10(a). Bohr’s Model for Hydrogen like Atoms:
1. 2 h mvr n 2. 2 2 2 18 1 2 2.178 10 2 / 13.6 2 n E z z E z J atom eV n n n 2 4 1 2 2 me E n 3. 2 2 2 2 2 0.529 Å 4 n n h n r Z e m Z 4. 2 6 2 2.18 10 / ze z v m s nh n
5.Revolutions per sec =
2 16 3 0.657 10 / 2 Z v r n
6.Time for one revolution =
16 3 2 1.52 10 2 r v/ n Z 7.En K E. .P E P E. .( . . 2 . .)K E 8. 2 2 0 1 . . 1/ 2 , . . 4 ze K E mv P E r
Illustration 1.
What is the principal quantum number of H atom orbital is the electron energy is –3.4 eV? Also report the angular momentum of electron.
Solution. E1 for H = –13.6 eV Now, n 21 E E n 2 13.6 3.4 n n = 2
Now, Angular momentum (mur) =
34 34 1 2 6.626 10 . 2.1 10 sec 2 2 3.14 h n J . Illustration 2.
Calculate the energy, velocity and radius of electron in Li2+ ion. Solution.
For Li2+ ion, z = 3, n = 1, then
radius = 2 1 0.529Å 0.529 Å 0.1763Å 3 n z
velocity = 2.18 108 cm/ sec z 3 2.18 108 cm/ sec . n 8 6.54 10 cm/ sec. Energy 2 2 9 13.6 13.6 1 z ev ev n 122.4 ev .
11. DEFINITION VALID FOR SINGLE ELECTRON SYSTEM : (i) Ground state:
lowest energy state of any atom or ion is called ground state of the atom. Ground state energy of H-atom = –13.6 eV
Ground state energy of He+ ion = –54.4 eV (ii) Excited state:
State of atom other than the ground state are called excited states:
n = 2 first exited state
n = 3 second exited state
n = 4 third exited state
n = n + 1 nth exited state
(iii) Ionisation energy (IE) :
Minimum energy required to move an electron from ground state to
n is called ionisation energy of the atom or ion.
Ionisation energy of H-atom = 13.6 eV Ionisation energy of He+ ion = 54.4 eV
Ionisation energy of Li+2 ion = 122.4 eV (iv) Ionisation Potential (I.P):
Potential difference through which a free electron must be accelerated from rest, such that its kinetic energy becomes equal to ionisation energy of the atom is called ionisation potential of the atom.
I.P. of H atom = 13.6 V I.P of He+ Ion = 54.4 V
(v) Excitation Energy:
Energy required to move an electron from ground state of the atom to any other state of the atom is called excitation energy of that state.
excitation energy of 2nd state = excitation energy of 1st state = 1st excitation energy = 10.2 eV. (vi) Excitation Potential:
Potential difference through which an electron must be accelerated from rest to so that its kinetic energy become equal to excitation energy of any state is called excitation potential of that state.
excitation potential of third state = excitation potential of second excitate state = seconds excitations potential = 12.09 v.
(vii) Binding Energy ‘or’ Separation Energy:
Energy required to move an electron from any state to n is called binding energy of that state. Binding energy of ground state = I.E. of atom or Ion.
Illustration.
A single electron system has ionisation energy 11180 kJ mol–1. Find the number of protons in the nucleus of the
system. Solution. I.E. = 2 19 2 21.69 10 Z J n 3 2 19 23 2 11180 10 21.69 10 6.023 10 Z I Z = 3 12. HYDROGEN SPECTRUM:
1. Study of Emission and Absorption Spectra:
An instrument used to separate the radiation of different wavelengths (or frequencies) is called spectroscope or a spectrograph. Photograph (or the pattern) of the emergent radiation recorded on the film is called a spectrogram of simply a spectrum of the given radiation. The branch or science dealing with the study of spectra is called
spectroscopy. Emission Spectra
When the radiation emitted from some source e.g. from the sun or by passing electric discharge through a gas at low pressure or by heating some substance to high temperature etc, is passed directly through the prism and then received on the photographic plate, the spectrum obtained is called ‘Emission spectrum’.
Depending upon the sources of radiation, the emission spectra are mainly of two types:
(i) Continuous spectra:
When white light from any source such as sun, a bulb or any hot glowing body is analysed by passing through a prism it is observed that it splits up into seven different wide bound of colours from violet to red. These colours are so continuous that each of them merges into the next. Hence the spectrum is called continuous spectrum.
Photographic Plate Prism Beam White light Slit V I B G Y O R
(ii) Line Spectra:
When some volatile salt (e.g., sodium chloride) is placed in the Bunsen flame or an electric discharge is passed through a gas at low pressure light emitted depends upon the nature of substance.
Photographic Prism Beam Slit Platinum wire 5896 Å
It is found that no continuous spectrum is obtained but some isolated coloured lines are obtained on the photographic plate separated from each other by dark spaces. This spectrum is called ‘Line emission spectrum’ or simply Line spectrum.
2. Absorption spectra:
When white light from any source is first passed through the solution or vapours of a chemical substance and then analysed by the spectroscope, it is observed that some dark lines are obtained in the otherwise continuous spectrum. These dark lines are supposed to result from the fact that when white light (containing radiations of many wavelengths) is passed through the chemical substance, radiation. of certain wavelengths are absorbed, depending upon the nature of the element.
Photographic Plate Prism NaCl Solution Slit V I B G Y} O R White light
Dark lines in yellow region of continuous spectrum Photographic Plate Prism Beam Slit
EMISSION SPECTRUM OF HYDROGEN: 12(a). H-Atom Spectrum
When an electric discharge is passed through hydrogen gas at low pressure, a bluish light is emitted. When a ray of this light is passed through a prism, discontinuous line spectrum of several isolated sharp lines is obtained as shown in figure. n= 6 5 4 3 2 n=1 Energy Lymann series Balmer series Paschen series Bracket series Pfund series
Energy levels of H-atom
All these lines observed in the hydrogen spectrum can be classified into the series as is tabulated in the table.
The hydrogen Spectrum
Region Spectral lines n1 n2
UV Lyman series 1 2,3,4,...
Visible Balmer series 2 3,4,5,...
I R Paschen series 3 4,5,6,...
far-I.R. P fund series 5 6,7,8,...
Illustration
Calculate the highest wavelength of line spectra of H-atom when the electron is situated in 3rd excited state. Solution.
Highest wavelength means lowest energy difference of electronic transition from one energy level to other energy level.
Hence, lowest energy transition will be n4to n .3
4 1 13.6 0.85 16 E ev ev 3 1 13.6 1.54 9 E ev ev 4 3 ( 0.85 1.54) E E E ev 19 0.69 0.69 1.6 10 hc ev J 34 8 7 19 6.626 10 3 10 18 10 0.69 1.6 10 m m 13. DE-BROGLIE RELATIONSHIP
According to de-Broglie matter has dual character i.e. wave as well as particle, if the wavelength of matter be having mass m moving with velocity v, then,
h mv
where h is Planck’s constant 34
(6.626 10 Joulsec).
wave in phase wave out of phase Case - I Case - II
For electron moving around a nucleus in a circular path, two possible waves of different wavelengths are possible.
In case I, the circumference of the electron orbit is an integral multiple of wavelength. In case II, wave is destroyed by interference and hence, does not exist.
Therefore, the necessary condition for a stable orbit of electron of radius r is, 2 r n
when n1, 2,3,etc. As h mv 2 , 2 nh nh r or mvr mv
This is simply the original Bohr condition for a stable orbit. Hence, the Bohr’s model of H-atom is justified by de-Broglie relationship.
13.(a)
De-Broglie Relations:
h h
de-Broglie pointed out that the same equation might be applid to material particle by suing m for the mass of the particle instead of the mass of photon and replacing c, the velocity of the photon, by v, the velocity of the particle.
2 ( . .)
h h
mv m K E
From the de-Broglie equation it follows that wavelength of a particle decreases with increase in velocity of the particle. Moreover, lighter particles would have longer wavelengths than heavier paticles, provided the velocity is equal.
If a charged particle Q is accelerated through potential difference V from rest then de-broglie wavelength is
2 h mQV
de-Broglie concept is more isgnificant for microscopic or sub-microscopic particles whose wavelength can be measured.
The circumference of the nth orbit is equal to n times the wavelength of the electron. 2rn n
Wavelength of electron is always calculated using De-broglie calculation.
Illustration. 1
Calculate the wavelength of a body of mass 1 kg moving with a velocity of 10 m sec–1. Solution.
We know, h
mv
Substituting the values, m = 1, mg = 10–6 kg, v = 10 m sec–1.
and h6.625 10 34kg m s2 1 34 29 6 6.625 10 6.625 10 10 10 m Illustration. 2
13.6 eV is needed for ionisation of a hydrogen atom. An electron in a hydrogen atom in its ground states absorbs 1.50 times as much energy as the minimum energy required for it to escape from the atom. What is the wavelength
of the emitted electron? ( 9.109 1031 , 1.602 1019 , 6.63 10 34 .
e
m kg e coulomb h J s
)
Solution.
1.5 times of 13.6 eV, i.e., 20.4 eV, is absorbed by the hydrogen atom out of which 6.8 eV (20.4 – 13.6) is converted to kinetic energy.
KE = 6.8 eV = 6.8 ( 19 1.602 10 coulomb) (1 volt) = 18 1.09 10 J. Now, 1 2 2 KE mv or, 18 6 31 2 2(1.09 10 ) 1.55 10 / . (9.109 10 ) KE J v m s m kg 34 10 31 6 (6.63 10 . ) 4.70 10 . (9.109 10 )(1.55 10 / ) h J s metres mv kg m s
14. HEISENBERG’S UNCERTAINTY PRINCIPLE (1927):
If subatomic particles have wave nature then we can’t pinpoint where exactly a particle is. The idea was defined by Heinsenberg as - There is a limit to the precision to which the position and momentum of a particle may be determined
simultaneously-/ 4
x p h
Where uncertainty in position of an electron.x p
uncertainty in momentum of an electron, i.e., when we try to determine position for a subatomic particle correctly, uncertainly in momentum will be very large or when we try to determine the momentum correctly then uncertainly in position will be large.
ORIGIN OF QUANTUM THEORY
When solid body heated it emit radiations in the forms of waves. The wave nature of light can be explained by diffraction interference etc. But some other observable properties such as photoelectric effect, compton effect could not be explained from wave nature. Hence a different theory is needed to explain these facts.
Illustration.
Calculate the uncertainty in the velocity of a wagon of mass 2000 kg whose position is known to an accuracy of 10 m
.
Solution.
Uncertainty in position, x 10m
Mass of the wagon, m2000kg
According to Heinsenberg’s principle, . 4 h x m v 34 2 1 39 1 6.626 10 2.64 10 4 . 4 3.14 2000 10 h kgm s v ms m x kg m .
14(a). Black Body Radiation
When radiation falls on an object, a part of it is reflected, a part is absorbed and the remaining part is transmitted because no object is a perfect absorber. But the black body (e.g., a metallic hollow sphere with a small hole, blackened on the inside surface) absorbs completely all the radiations falls on it by successive reflections inside the enclosure. T1 T2 T3 Wave length energy T1>T2>T3
The black body is not only a perfect absorber of radiation energy, but also an ideal radiation, i.e., when the black body is heated, it radiates the maximum amount of energy. The energy which is radiated is dependent on the temperature of the black body and is independent of the nature of the interior material.
The curves represent the distribution of radiation from a black body at different temperatures. The shape of the curves couldn’t be explained on the basis of classical electromagnetic theory in which it was assumed that the body radiates energy continuously. So, the intensity of radiation should increase continuously without limits as the frequency increases. But the experimental observations are contrary to the classical view. For each temperature, there is a maximum in the curve corresponding to a particular wavelength, indicating the maximum radiation of energy. At higher temperature, the position of the maximum in the curve shifts towards shorter wavelength and becomes more pronounced. To explain this black body radiation, Max Planck put forward new quantum theory.
Planck Quantum Theory
(a) Radiation energy is not emitted or absorbed continuously but discontinuously in the form of tiny bundles of energy, called quanta.
(b) Each quanta is associated with a defined amount of energy (E) which proportional to the frequency of
radiation i.e, Ev or, Ehv where h is Planck’s constant 34
(6.626 10 Jsec).
(c) A body can emit or absorb energy only in whole number multiples of quanta i.e. Enhv where n1, 2,3etc.
15. PHOTOELECTRIC EFFECT
Sir J.J. Thomson has discovered this phenomenon of ejection of electron from the surface of a metal when light of suitable frequency of strikes on it.
Only few metals show this effect under the action of visible light, but many more show it under the action of more energetic u.v. light. For every metal, there is a minimum frequency of incident radiation necessary to eject electron from that metal surface, is known as Threshold frequency (v ). This 0 v varies metal to metal.0
The number of ejected electrons from the metal surface depends upon the intensity of the incident radiation. Greater the intensity, the larger is the number of ejected electrons.
Hence, according to quantum theory, when a photon of light of frequency (vv0) strikes on an electron in a
metal, it imparts it entire energy to the electron. Then some of its energy (equal to binding energy of electron with the nucleus) is consumed to separate the electron from the metal and the remaining energy will be imparted to the ejected electron.
2 0
1 2
hvhv mv where hv is the binding energy of work function of the electron and 0 2
1
2mv is the kinetic energy of electron. Alkali metals are mainly used for photoelectric effect. Cesium, amongst alkali metals, has lowest threshold energy and used largely in photoelectric cell.
16. SHAPES OF ORBITALS
s-orbital: they do not have directional character. They are spherically symmetrical. The s-orbital of higher
energy levels are also spherically symmetrical. They are more diffused and have spherical shells within them where probability of finding the electron is zero.
z
x y
Node
2s orbital
In the s-orbital, number of nodes is (n – 1)
p-orbital: ‘p’ orbital has a dumb-bell shape and it has a directional character.
The two lobes of a p-orbital are separated by a plane that contains the nucleus and is perpendicular to the corresponding axis. Such plane is called a nodal plane because there is no probability of finding the electron.
+ px z y + py z y x x + pz z y x
In the absence of an external electric or magnetic field, the three p-orbitals of a particular energy level have same energy and are degenerate. In the presence of an external magnetic field or electric field this degeneracy is removed.
d-orbitals: For d-orbitals five orientations are possible viz., dxy, dyz, dxz, dx2y2, dz2. All these five orbitals in the absence of magnetic field are equivalent in energy and are degenerate.
dxy x y dxz x z dyz x z
These three ‘d’ orbitals are similar. The maximum probability of finding the electron is in lobes which are directed in between the axes. Nodal region is along the axes.
dz2
z
dx2-y2
y
x
These two d-orbitals are similar. Probability of finding the electron is maximum along the axes and the nodal region is in between the axes.
17. QUANTUM NUMBERS
These are used to determine the region of probability of finding a particular electron in an atom.
(a) Principal quantum number (n):
This denotes the energy level or the principal or main shell to which an electron belongs. It can have only integral values 1, 2, 3 etc. The letter K, L, M ... are also used to designate the value of n. Thus, an electron in the K shell has n = 1, that is L shell has n = 2 and so on.
Illustration. 1
The principal quantum number of 2s-electron is
Solution.
n = 2
(b) Azimuthal quantum numbers (l):
This denotes the orbital (Sub-level) to which an electron belongs. It gives an idea about the shape of the orbital.
l can have any value from 0 to (n – 1), for a given value of n,
i.e. l = 0, 1, 2, ... (n – 1)
Value of l 0 1 2 3
Sub Shell s p d f
The value of orbital angular momentum of the electron for a given value of ‘l’ is ( 1) 2
h l l
(c) Magnetic quantum number (m):
It gives us the idea about the orientations an orbital can have in space in the presence of magnetic field. The values of ‘m’ depends on ‘l’ orbital quantum number.
Total value of m = (2l + 1) and it varies –l to +l.
For example, for l = 0 the value of magnetic quantum number m is also equal to zero, i.e. ‘s’-orbital can have only one orientation in space in presence of magnetic field.
(d) Spin quantum number (s):
clockwise direction or in an anticlockwise direction. Its value is 1 2 or 1 2 corresponding to clockwise or anticlockwise spin.
(1) The value of spin angular momentum for a given value of s is ( 1) 2
h s s
(2) The spin magnetic moment of electron (excluding orbital magnetic momentum) is given by
( 2)
effective n n BM
(Where n = Number of unpaired electrons).
18. DISTRIBUTION OF ELECTRONS IN AN ATOM
The filling up of orbitals with electrons takes place according to certain rules which are given below:
(i) The maximum number of electrons in a main shell is equal to 2n2, where n is the principal quantum number.
(ii) The maximum number of electrons in a sub-shell like s, p, d, f is equal to (2l + 1), where l is the azimuthal quantum number for the respective orbitals. Thus s, p, d, f can have a maximum of 2, 6, 10 and 14 electrons respectively.
(a) Afbau Principle
According to the principle, “Electrons are added progressively to the various orbitals in the order of increasing energy”.
What does the word ‘Afbau’ mean?
Afbau is a German term which means building up or construction. The energy of various orbitals increase in the order given below:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p < 8s < ...
(i) A new electron enters the orbitals for which (n + l) is minimum, e.g. if we consider 3d and 4s orbitals, the electron will first enter 4s-orbitals in preference to 3d.
This is because the value of (n + l) for 4s-orbitals is less (4 + 0 = 4) than that for 3d-orbital (3 + 2 = 5) (ii) In case where (n + l) values are the same, the new electron enters the orbitals for which ‘n’ is minimum, e.g. in a choice between 3d and 4p for which (n + l) values are same (3 + 2 = 5, 4 + 1 = 5), the electron will prefer to go to the 3d-orbital, since n is lower for this orbital.
(b) Pauli’s Exclusion Principle
It states that it is impossible for two electrons in a given atom to have same set of quantum numbers.
Example: (a) n = 2, l = 0, m = 0, s = +1/2 n = 2, l = 0, m = 0, s = –1/2 (b) n = 2, l = 1, m = 0, s = +1/2 n = 2, l = 1, m = 0, s = –1.2 n = 2, l = 1, m = +1, s = +1/2 n = 2, l = 1, m = +1, s = –1/2 n = 2, l = 1, m = –1, s = +1/2 n = 2, l = 1, m = –1, s = –1/2
(c) Hund’s Rule of Maximum Multiplicity
give maximum number of unpaired electrons. In other words it means that pairing begins with the introduction of the second electron in the s-orbital, the fourth in p, etc.
What is the electronic configuration of Cu (Z = 29)? 1s2 2s2 2p6 3s2 3p6 3d10 4s1
Exceptional Electronic Configuration
Some atoms such as copper and chromium exhibit exceptional electronic configuration. For example:
Cr(Z = 24) has an electronic configuration 1s2 2s2 2p6 3s2 3p6 3d5 4s1
It is because of the extra stability associated with the half-filled and completely filled orbitals.
CONCEPT BUILDING EXAMPLES
Example 1.
Find out the energy of the electron in the first excited state in an H-atom.
Solution.
Energy of an electron in H-like atom is given by
2 8 2 (2.18 10 ) Z ; E J n
where z is the number of protons and n is number of shell in which electron is present.
For the first excited state, n 2
2 18 2 1 (2.18 10 ) 2 E J 19 5.45 10 E J Example 2.
The K.E. of a moving electron is 5 10 25J ; Calculate its velocity and the wavelength. Solution : 2 1 . . 2 K E mv 25 3 1 31 2( . ) 2 5 10 1.048 10 9.1 10 K E J velocity v ms m kg Now, wavelength h mv 34 7 31 3 1 6.626 10 6.949 10 (9.1 10 ) (1.048 10 ) Js m kg ms Example 3.
If the electron of the hydrogen atom has been excited to a level corresponding to 10.2 electron volts, what is the wavelength of the line emitted when the atom returns to its ground state?
Solution. 2 1 E E E E hv hc E 10.2 E eV E10.2 1.6 10 19J
34 8 19 6.624 10 3 10 10.2 1.6 10 m 34 8 9 19 6.624 10 3 10 10 10.2 1.6 10 n m 121.8nm . Example 4.
An electron jumps from fourth excited state to the ground stable in Li+2 -ion and the energy released in the form
of photon is allowed to strike a metal (x) surface whose work function ( ) is 1 10 18J
. What is the K.E. & velocity of the electron ejected.
Solution.
The amount of energy released is given by
18 2 2 2 1 2 1 1 (2.18 10 ) E J z n n
where z is the atomic number of H-like atom.
18 2 1 18 (2.18 10 ) (3) 1 1.23 10 16 E J J Now, hvK E. . 18 18 . . (1.23 10 1 10 ) K E J 2 18 1 2.3 10 2mv J 19 1 5 1 31 2 2.3 10 7.11 10 9.1 10 v ms ms Example 5.
Ionisation energy of hydrogen atom is 13.6 eV. What will be the ionisation energy of He+ and Li++ ions? Solution.
As we know ionisation energy for one electron system is given by
2 2 . . 13.6 z I P eV n
For hydrogen atom, z = 1, n = 1, So, I.P = 13.6 eV For He+ ion z = 2 and n = 1.
So, I.P = 13.6 13.6(2)2
i.e., . .I P 54.4eV For Li2+, z = 3 & n = 1; I.P. = 13.6 (3) 2
i.e., .I P122.4eV .
Example 6.
A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.2 eV and 17 eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energy 4.25 eV and 5.95 eV respectively. Determine the values of
n and z (ionisation energy of hydrogen atom = 13.6 eV).
Solution.
Total energy liberated during transition of electron from nth shell to first excited state (i.e., 2nd shell)
10.2 17 17.2 eV
12
27.2 1.602 10 erg
2 2 2 1 1 2 H hc R Z hc n 12 2 2 2 1 1 27.2 1.602 10 2 H R Z hc n ... (i)
Similarly, total energy liberated during transition of electron from nth shell to second excited state (i.e. 3rd shell):
4.25 5.95 10.2eV 12 10.2 1.602 10 erg 12 2 2 2 1 1 10.2 1.602 10 3 H R Z hc n ... (ii)
Dividing eq. (i) by eq. (ii) n 6
On substituting the value of n in eqns. (i) or (ii) Z .3
Example 7.
1 mol of He+ ion excited. Spectral analysis showed the existence of 50% ions in 3rd level, 25% in 2nd level and
remaining 25% in ground state. Ionisation energy of He+ is 54.4 eV; calculate total energy evolved when all the
ions return to ground state.
Solution.
25% of He+ ions are already in ground state, hence energy emitted will be from the ions present in 3rd level and
2nd level. 2 2 2 1 2 1 1 ( ) E IP n n
per ion or atom.
0 2 2 3 1 1 1 (54.4) 2 1 3 N E for 0 2 Nions falling to ground state
0 4 54.4 9 N eV and
2 1 (54.4) 0 12 12 4 1 2 N E for 0 4 Nions falling to ground state.
0 3 54.4 16 N eV
Hence total energy = 0
4 3 54.4 9 16 N 23 91 54.4 6.023 10 144eV 23 91 19 54.4 6.023 10 1.6 10 144 J 4 331.13 10 J Example 8.
Estimate the difference in energy between the 1st and 2nd Bohr orbit for a hydrogen atom. At what minimum atomic number, would a transition from n = 2 to n = 1 energy level result in the emission of X-rays with
8
3.0 10 m
. Which hydrogen atom like species does this atomic number correspond to?
Solution.
For hydrogen atom, the expression for energy difference between two energy level is,
2 2 2 2 1 1 H E R hc n n
So, 2 1 7 34 8 1 1.09677 10 6.626 10 3 10 1 4 E E E 8 1.635 10 J
For hydrogen like species, the same expression is
2 2 2 2 2 1 1 1 H z R n n or, 2 2 2 1 2 1 1 1 H z R n n So, 8 2 7 1 3 (1.09677 10 ) 4 3 10 z or, z = 2
So, the species is He+ because z = 2. Example 9.
What is the degeneracy of the level of the hydrogen atom that has the energy
(a) –RH, (b) –RH/9 Solution. 2 / n H E R n (a) E RH, n 1 when n ,1 l 0, m e 0
The level is degenerate.
(b) E RH/ 9
n 3 when n ,3 l 0,1, 2
when l 0, m e 0 (3s-orbital)
when l 1, m e 1, 0,1 (3p-orbital)
when l 2, me 2, 1, 0,1, 2 (3d-orbitals)
This is 1 + 3 + 5 = 9 states in all. The degeneracy is 9.
Example 10.
Calculate the uncertainty in the velocity of a ball of mass 150g if uncertainty in this position is
34
1Å (h6.6 10 Js)
Solution.
From Heisenberg uncertainty principle we know that
4 4 4 h h h x p or xm v or v m x Substituting value: m150g0.15kg, x 1Å1010m 34 10 6.6 10 4 3.14 10 0.15 v 24 1 3.499 10 v ms .
Subjective Solved Examples Exampe1. 11
Calculate the waelength and wave number of the spectral line when an electron in H-atom falls from higher energy state n 3 to a state n . Also determine the energy of a photon to ionise this atom by removing the2 electron from 2nd Bohr’s orbit. Compare it with the energy of photon required to ionise the atom by removing the electron from the ground state.
+e n=2 n=3 Photon emitted +1e n=2 n= Photon absorbed n=1
First calculate the energy (E) between the To ionise the atom from n 2, the
Bohr orbits n 3 and n 2 using: responsible transition will be n2n .
18 2 2 2 1 2 1 1 2.18 10 E Z J n n 18 2 (2 ) 2 2 1 1 2.18 10 1 2 E J 18 2 (3 2) 2 2 1 1 2.18 10 (1) ( 1) 2 3 E J Z 19 5.45 10 J 19 3.03 10 J
To ionise the atom from ground state
Now this energy difference is the energy of the (n 1), the transition is 1 .
photon emitted. 18 2 2 2 1 1 2.18 10 1 1 E Photon hc E hv hcv 2.18 10 18 J hc 3.03 10 19 6560.3Å and v 1 1.52 106m1 Example 12
A hydrogen atom in the ground state is hit by a photon by a photon exciting the electron to 3rd excited state. The electron then drops to 2nd Bohr orbit. What is the frequency of radiation emitted and absorbed in the process?
Solution.
Energy is absorbed when electron moves from ground state (n 1) to 3rd excited state (n 4).
+e n=2 Photon absorbed n=1 n=1 n=4
First calcuate the energy difference btween n and 1 n 4.
Use: 18 2 (1 4) 2 2 1 2 1 1 2.18 10 E Z n n .
Here, Z1,n11,n2 4 Put n 1 2 and n 2 4 in the expression of E , to get: (1 4) 18 2 2 2 1 1 2.18 10 1 1 4 E J 18 2 (4 2) 2 2 1 1 2.18 10 1 2 4 E J 18 2.04 10 J 4.08 10 19 J
This is the energy of the photon absorbed. This is the energy of the photon emitted.
Use: 2.04 10 18 Photon E hv J to get: Use: 4.08 10 19 Photon E hv J v3.08 10 15 Hz v6.16 10 14 Hz
Similarly, when electron jumps from 4
n to n 2, energy is emitted and is given by the same relation.
Example 13
A hydrogen like ion, He Z( 2)
is exposed to electromagnetic waves of 256.4 Å. The excited electron gives out induced radiations. Find the waelength of the indicated radiations, when electron de-excites back to the ground state. R = 109677 cm–1.
Solution:
He+ ion contains only one electron, so Bohr’s From
3
n , the electron can fall back to the medhot is applicable here. It absorbs a photon of ground state in three possible ways (transitions)
wavelength 256.4Å. Assume the electron to 31, 32, 21
be in ground state initially. Let it jumps to an Hence three possible radiations are emitted.
excited state n2. Find the wavelengths corresponding to these
2 2 2 1 2 1 1 1 v R Z n n transitions. Substitute for 256.4Å 256.4 108 cm
, The wavelength ( ) for transition, 3 will1 be same i.e., 256.4 Å. Find for 32 and
2 using the same relation.1
109677, 2 R Z for He+ ion, (31)256.4 ,Å (32)1641.3Å 1 1 n and find n .2 (21)303.9Å 7 2 8 2 2 1 2 1 1 1 109677 10 (2) 256.4 10 1 n n 2 3 n=3 n=2 n=1 Example 14
Hydrogen gas when subjected to photon-dissociation, yields one normal atom and one atom possessing 1.97 eV more energy than normal atom. The bond dissociation energy of hydrogen moleucle into normal atoms is 103 kcals mol–1. Compute the wave length of effective photon for photon dissociation of hydrogen molecule in
the given case.
Solution: 2
H HH
where H is normal H -atom and H is excited H-atom. So the energy requird to dissociate * H in this2
manner will be greater than the usual bond energy of H molecule.2
E (absorbed) = dissociation energy of H extra energy of excited atom.2
3
103 10 cal per mol
(given) 3 19 23 103 10 4.18 7.17 10 / 6 10 J atom
The extra energy possessed by excited atom is 1.97 eV
19 19 1.97 1.6 10 J 3.15 10 J E (absorbed) = 7.175 1019 3.15 1019J 18 1.03 10 J
Now calculate the wavelength of photon corresponding to this energy.
18 1.03 10 1930 Å Photon hc E Example 15
An electron in the first excited state of H-atom absorbs a photon and is further excited. The de Broglie wavelength of the electron in this state is found to 13.4 Å. find the wavelength of photon absorbed by the electron in Å. Also find the longest and shortest wavelength emitted when this electron de-excites back to ground state.
Solution:
Note: The energy state n is known as Ground State1 The energy state n 2 is known as First Excited State The energy state n 3 is known as Second excited State and so on.
+e
n=2 n=n
Photon
The electron from n 2 absorbs a photon and is further excited to a higher energy level (let us say n). The electron in this energy level (n) has a de Broglie wavelength ( ) 13.4Å
e e e h m v and 2.18 106 1 n Z v ms n [v n is the velocity of e – in nth Bohr orbit] v h 2.18 106 1 m n 34 10 31 6.626 10 (13.4 10 ) (9.1 10 ) 6 1 2.18 10 n n 4
Now find the wavelength of the photon responsible for the excitation from n2to n4. Using the relation :
18 2 2 2 (2 4) 1 2 1 1 2.18 10 E Z n n 19 4.09 10 J [n1=2, n2=4, Z=2] 19 (2 4) 4.09 10 hc E 4863.1Å
The Longest wavelength emitted when this electron E(4 3) 1.06 1019
(from n 4) falls back to the ground state will corresponds Photon 1.06 1019 hc E E J
to the minimum energy transition. 18752.8Å
The transition corresponding to minimum energy will Shortest wavelength : 41
be 4 .3 (4 1) 18 2 2 2 1 1 2.18 10 1 1 4 E
Note: The transition corresponding to maximum energy 2.04 10 18 J
will be 4 .1 (4 1) Photon hc E E
(Energy diff.) Photon
hc E E hv 973.2Å 1 Photon Photon E or E v
Using the same relation : (4 3) 18 2 2 2
1 2 1 1 2.18 10 E Z n n [n13,n2 4,Z 2] Example 16
A single electorn orbits around a stationary nucleus of charge Ze , where Z is a constant and e is the magnitude of electronic charge. It requires 47.2 eV to excite the electron from second Bohr orbit to the third Bohr. Find :
(a) the value of Z
(b) the energy required to excite the electron from n3to n .4
(c) the wavelength of radiation required to remove electron from 2nd Bohr’s orbit to infinity (d) the kinetic energy, potential energy and angular momentum of the electron in the first orbit. (e) the ionisation energy of above one electron system in eV.
Solution.
Since the nucleus has a charge Ze , the atomic number of the ion is ‘ Z ’.
(a) The transition is n12n2 by absorbing a photon of energy 47.2 eV .3
E 47.2eV
Using the relation:
2 2 2 1 2 1 1 13.6 E Z eV n n 2 2 2 1 1 47.2 13.6 5 2 3 Z Z
(b) The required transition is n1 3 n2 by absorbing a photon of energy E4 .
Find E by using the relation:
2 2 2 1 2 1 1 13.6 E Z eV n n 2 2 2 1 1 13.6(5) 3 4 E eV E 16.53eV
Find E by using the relation: 2 2 2 1 1 13.6(5) 85 2 E E eV
Find of radiation corresponding to energy 85 eV.. 34 8 19 6.626 10 3 10 85 (1.6 10 ) hc E 146.16 Å
(d) If energy of electron be En, then KE = –En and PE = 2En
2 2 2 2 13.6 13.6 5 340 1 n Z E eV n (340 ) 340 KE eV eV 2( 340 ) 680 PE eV eV Angular momentum ( ) 2 h l n 34 6.626 10 1 2 l 34 1.05 10 J s
(e) The ionisation energy (IE) is the energy required to remove the electron from ground state to infinity. So the required transition is 1 . The ionisation energy
2 1 (IE) E 13.6( )Z eV 2 13.6 5 340 IE eV Example 17
With what velocity should an alpha ( ) particle travel towards the nucleus of a copper atom so as to arrive at a distance 10–13 m from the nucleus of the copper atom?
Solution.
As -particle appraoches towards the Cu nucleus, it decelerates due to repulsion from it and finally its velocity will become zero at point A (which is the turning point). After that, particle will move in the left direction (and accelerating)
A V=0 Cu Nucleus (+29e) r0 -particle V
To arrive at a distance (r0) from the nucleus, the kineticc energy of alpha particle should be equal to the electrostaic potential energy of it, i.e., KE = EPE
2 0 1 2 N Kq q m v r : m mass of -particle = 27 4 (1.67 10 kg) : v velocity of -particle = ? 9 2 9 10 / K Nm C
q charge on -particle =
19
2 ( 1.6 10 C) r0 = Minimum distance of approach
Note: q2 ,e m 4mp [ -particle is He2+ ion or He Nucleus]
qN = charge in Cu nucleus = Ze = 29(+1.6 10–19C)
d = distance from nucleus = 10–13 m
0 2Kq qN v m r
Substituting the given values, we get,
6
6.325 10 / .
v m s
Note: This is a simple cases where velocity of -particle is directed towards the centre of the Cu-nucleus.
A Cu
r0
-particle
V=0
Note: When there is a difference between the velocity vector of -particle and the Cu(target) nucleus, the trajectory is more complicated.
Target -particle
Example 18
Find the energy required to excite 1.22 litre of hydrogen atoms gas at 1.0 atm and 298 K to the first excited state of atomic hydrogen. The energy requied for the dissociatio of H-H bonds is 436 kJ/mol. Also calculate the minimum frequency of a photon to break this bond.
Solution.
Let us, first find the number of moles of hydrogne atoms.
2 1 1.22 0.05 0.0821 298 H PV n RT
Thus the energy required to break 0.05 moles of H2 (H-H bond) = 0.05 436 19.62 kJ.
Now calculate the energy needed to excite the H-atoms to first excited state i.e., to n 2 (First excited state is referred to n 2). 18 2 2 2 1 1 2.18 10 (1) / 1 2 E J atom 18 1.635 10 J atom/
No. o H atoms = (No. of H2 molecules)2
23 22
(0.05 6.02 10 ) 2 6.02 10
The energy required to excited the given number of H-atom = 6.02 1022 1.635 1018J 98.43kJ
So the total energy required 19.62 98.43 118.05 kJ
Now the energy required to break to single H-H bond = 3 19 23 436 10 7.238 10 6.023 10
= Energy supplied by the photon 7.328 10 19hv6.626 10 34( )v
v1.09 10 15 Hz
Example 19
Estimate the differnce in energy between 1st and 2nd Bohr’s orbit for a H-atom. At what minimum atomic number (Z), a transition from n to 2 n energy level would result in the emission of radiatio with wavelength1
8
3.0 10 m?
Which hydrogen atom like species this atomic number corresponds to? How much ionisation
potential is needed to ionise this species? 7 1
(R1.097 10 m)
Solution.
The difference in energy is given by E :
18 2 2 2 1 1 2.18 10 (1) / 1 2 E J atom 18 11 1.65 10 J 1.65 10 ergs 10.2eV
For a H-like atom, 3.0 10 m8
. (2 1) 18 2 2 2 1 1 2.18 10 1 2 E Z J Photon hc E Solve to get : Z = 2
Hence the H-like atom is He+ ion.
To ionise, He+ ion, ionisation energy (IE) = –(E 1) 2
( 13.6 2 ) 54.4
IE eV
The ionisation potential (IP) is the voltage difference required to generate this much energy.
IEqVe IP( )54.4eV
IP(required)54.4Volt
Example 20
A stationary He+ ion emits a photon correspondings to the first line (H )
of Lyman series. The photon thus
emitted, strikes a H-atom in the ground state. Find the velocity of the photoelectrons ejected out of the hydrogen
atom. The value of 7 1
1.097 10 .
R m
Solution.
The difference in energy (E) will be equal to the energy of the photon emitted. First line in Lyman series corresponds to the transition 2 .1
18 2 2 2 1 1 2.18 10 (2) / 1 2 E J atom
