NUCLEAR PHYSICS: (Father of nuclear Physics: Ernest Rutherford): The branch of Physics dealing with the study of atomic nucleus is called Nuclear Physics .It includes the study of properties of nucleus, nuclear phenomena, interaction of nuclei, etc. It is divided into 1) Particle Physics & 2) High energy Physics.
Nucleus: Discovered by Ernest Rutherford (New Zealand) & his co-worker Geiger & Marsden in 1911 from alpha particle scattering experiment.
Atomic number (Z): The number of protons in the nucleus of an atom is called atomic number. Number of electrons, number of protons and atomic number of an atom are always equal.
Mass number ( or Atomic mass( A): The total number of protons and neutrons present inside the nucleus of an atom is called mass number.
CONSTITUENTS OF NUCLEUS:
a) All the atomic nuclei are made up of elementary particles called protons & neutrons (except 1H1, no neutron)
b) Proton is positively charged particles whose charge e=1.6x10-19C & discovered by Ernest Rutherford in 1919. It is a stable particles
c) Neutron is chargeless particle having rest mass 1.6748x10-27Kg .It is unstable in Free State. It is discovered by James Chadwick in 1932.
d) Atom consists of equal number of protons & electrons. So, it is called electrically neutral.
e) A nucleus is positively charged because it contains proton which has positively charged( neutron - chargeless)
f) The nucleus consists of protons & neutrons, collectively called Nucleons
g) The Species of nucleus is known as nuclide.
h) A= Atomic mass or mass number, Z= Atomic number , N= number of neutron N=A-Z
GENERAL PROPERTIES OF NUCLEUS:
a) Nuclear size: The size of nucleus is very small. According to Rutherford, the size of nucleus (radius) is of order 10-14 m to 10-15 m while that of an atom is about 10-10m . The empirical formula for the nuclear radius is given as R=r0(A)1/3 where A is atomic mass & radius constant r0=1.2 fm =1.2 x 10-15m (experimentally determined constant)
b) Nuclear charge: A nucleus contains protons & neutrons .Protons is positively charged but neutrons are chargeless. So the nuclear charge is positive which is due to the charge carried by protons present in nucleus.
Nuclear charge (q) = +Ze where e=1.6x10-19C & Z= atomic number =number of protons.
c) Nuclear mass: It is the sum of masses of protons & neutrons present in a nucleus.
Nuclear mass (MN)=Zmp +Nmn where mn= mass of neutron mp=mass of proton
MN=mass of nucleus or nucleon
d) Nuclear density: The mass per unit volume of a nucleus is called nuclear density.ρ=𝒏𝒖𝒄𝒍𝒆𝒂𝒓 𝒎𝒂𝒔𝒔/𝒏𝒖𝒄𝒍𝒆𝒂𝒓 𝒗𝒐𝒍𝒖𝒎𝒆 =𝑨𝑴𝑵/4/3𝝅𝑹𝟑 =𝟑𝑨𝑴𝑵/𝟒𝝅𝑹𝟑=𝟑𝑨𝑴𝑵/(𝟒𝝅 𝒓𝟎 𝑨𝟏𝒍𝟑)𝟑=𝟑𝑴𝑵/𝟒𝝅𝒓𝟎^3
where MN= 1.67x10-27Kg and r0=1.2x10-15m
Putting values we get, ρ= 2.3x1017Kg/𝒎𝟑
This shows that density of nucleus is independent of mass number (A) and its value is very high. So, all nuclei have approx. same density. The nuclear density is not uniform throughout the nucleus. It has maximum value at the centre and decreases gradually as we move away from the centre of the nucleus.
Types of Nucleus: Nuclei is classified on the basis of number of protons, neutrons & radioactive character:
a) Isotopes: The nuclei having same atomic number but different mass number are called Isotopes.eg 1H1, 1 H2, 1H3
b) Isobars: The nuclei having same mass number but different atomic number are called Isobars. eg 6C14 & 7N14
c) Isotones: The nuclei having same number of neutron but different mass number & atomic number are called Isotones. eg 3Li7 & 4Be8
d) Mirror nuclei: The nuclei having same mass number but proton & neutron number interchanged are called mirror nuclei. eg 4Be7 (Z=4 , N=3) & 3Li7 (Z=3,N=4)
e) Isomers: The nuclei having same atomic number & mass number but differ in their radioactive characters are called Isomer. These nuclei are distinguish by their life time.
Einstein's mass-energy relation: In 1905, Einstein explained the interrelationship between mass & energy in his special theory of relativity. According to this theory, mass & energy are interchangeable i.e. mass can be converted into energy & vice-versa.
According to Einstein, the energy equivalent to a mass(∆𝑚) is, E=∆𝒎𝒄𝟐,
where c= 3x108m/s, is speed of light.
This equation represents Einstein's mass-energy relation. This relation explains the unification law of conservation of mass & law of conservation of energy into a single law called law of mass-energy conservation. So, this relation is also known as mass-energy equivalence.
According to this relation, 1Kg of mass of any material is equivalent to 9x 1016J of energy which is equal to 2.5x1010KWH (i.e. kilowatt-hour).
The mass of a body is not constant quantity. It varies with the speed of a body.
If we give energy (E) to the matter then its mass will increase. The increase in mass (∆𝑚) is given by ∆𝒎 =𝑬/𝒄^2 ,where mass is expressed in 𝑴𝒆𝑽/𝑪𝟐
Einstein's mass energy relation E=mc2 does not tell us that the mass travels with the speed of light(c). Here c2 is a conversion factor which is used to convert mass into energy & vice-versa.
Verification of E=mc2:
According to Newton’s second law of motion, force acting on a body is defined as the rate of change of momentum.
i.e. F= dp/dt =d(mv)/dt =mdv/dt +vdm/dt…………(1)
If this force F displaces the body ny a distance dx, its energy increases by,
dK = F dx = {mdv/dt +vdm/dt }dx = mdv/dt dx+vdm/dt dx =m dv dx/dt+v dm dx/dt = m v dv + v2 dm
∴ dK = mv dv + v2dm, ……………..(2)
According to Einstein’s relation of relativistic mass or Mass -velocity relation:
m =m0/√𝟏−𝒗𝟐/𝒄𝟐……(3)
squaring and manipulating (3): we get
m2c2 –m2v2 = 𝑚02c2 ……….(4)
Again, differentiating (4)(with m0 and c are constant): then we get
c2 2m dm – m2 2v dv – v2 2m dm = 0
or, c2 dm = mv dv + v2 dm ……..(5)
From (2) and (5): dK = c2 dm ………….(6)
If particle is accelerated from rest to a velocity v, let its mass m0 increases to m.
Now, integrating (6); k ∫ dK=C^2∫ m dm
0. m0
K= c2 (m-m0)
K = c2(m – m0)
or, K + m0c2 = mc2 ………(7)
Here, m0c2 is the energy associated with the rest mass of the body and K is the kinetic energy. Thus, the total energy of the body is given by E=mc2 ……..(8)
This is Einstein’s mass- energy equivalence relation. Hence, verified
Significance of Einstein's mass-energy equivalence:
a) It gives a relationship between mass & energy. Thus, it shows that mass & energy
can be converted into each other.
b) It forms the basis of understanding nuclear reactions like Fission & Fusion.
c) Mass -velocity relation: m=m0√𝟏−𝒗𝟐/𝒄𝟐.
d) The conversion of mass into energy can be seen in many devices like Atom Bomb, hydrogen Bomb etc but yet the Scientist has not devised a machine that can convert energy into mass.
Note: The energy equivalent of mass of an electron, proton & neutron are respectively given by,
me= 0.511MeV mp = 938.28 MeV mn=939.57 MeV
ONE ELECTRON VOLT (1eV): It is defined as the kinetic energy gained by a particle with charge equal to that of an electron or a proton when accelerated through a potential difference of 1 volt. Thus, by work-energy , we have 1eV= charge X potential difference = (1e)X (1 Volt) = 1.6x 10-19 X(1Volt)
=1.6x10-19J
So, 1eV= 1.6x10-19J
Also, 1MeV= 1.6x10-13J
ATOMIC MASS UNIT (amu or u):
One atomic mass unit (1 amu) is defined as the (𝟏/𝟏𝟐)th of mass of one 6C12 atom. As we know,
12 gram of 6C12 contains 6.023x1023
Now, mass of 6.023x1023 atoms of 6C12 = 12 gram
So, mass of 1 atom of 6C12 =𝟏𝟐/𝟔.𝟎𝟐𝟑𝐱𝟏𝟎𝟐𝟑 gram
By definition, 1amu = 1/12[𝟏𝟐/𝟔.𝟎𝟐𝟑𝐱𝟏𝟎𝟐𝟑 ] gram =1.66x10-24 Kg
Hence, 1 amu = 1.66x10-27 Kg
RELATION BETWEEN amu & MeV:
Since, 1 amu = 1.66x10-27 Kg
We know from Einstein's mass-energy relation, the equivalent energy of this mass isE=mc2 = (1.66x10-27)x(3x108)2 =1.494x10-10J
But, 1eV= 1.6 x10-19J
Therefore, Energy equivalence of 1 amu = 𝟏.𝟒𝟗𝟒𝐱𝟏𝟎−𝟏𝟎/𝟏.𝟔𝒙𝟏𝟎−𝟏𝟗 J = 931MeV
MASS DEFECT (∆𝐦): The difference between the sum of the masses of constituent Nucleon & mass of a nucleus is called the Mass defect.
The total mass of all the constituent Nucleon is always greater than the mass of the nucleus.
Mass defect = sum of masses of the constituent nucleon - Rest mass of nucleus
∆𝐦 =(Zmp+Nmn) - M
∆𝐦 =[Zmp+(A-Z)mn]- M
where, Z = atomic number mp = mass of proton mn = mass of
neutron M= rest mass of nucleus
PACKING FRACTION (PF): The mass defect per nucleon is called Packing Fraction. It is
the measure of comparative stability of an atom.
Packing fraction (PF) = 𝐦𝐚𝐬𝐬 𝐝𝐞𝐟𝐞𝐜𝐭(∆𝐦)
𝐦𝐚𝐬𝐬 𝐧𝐮𝐦𝐛𝐞𝐫(𝐀)
a) If PF > 0 then the nucleus is not stable.
b) If PF < 0 then nucleus is more stable
BINDING ENERGY or NUCLEAR BINDING ENERGY: The Binding energy of a nucleus is
defined as the minimum energy required to separate the nucleon from the nucleus for
apart to form stable nucleus.
It is also defined as the minimum energy required to bind the nucleons to form a stable
nucleus. It is equivalent to mass defect. Thus mass defect is a measure of the Binding
energy of the nucleus.
∆𝐦 = (𝐙𝐦𝐩 + 𝐍𝐦𝐧) − 𝐌 ∆……………..(1)
If ∆𝐦 is in Kg then Binding energy of nucleus is given by
EB=(∆𝐦 ) c
2
EB= [(𝐙𝐦𝐩 + 𝐍𝐦𝐧) − 𝐌 ] c
2 ………………..(2)
If ∆𝑚 is in amu or u , then Binding energy of nucleus is given by
EB=(∆𝐦 ) X 931 Mev ………………..(3)
a) if EB> 0 then the nucleus is stable & external energy must be supplied to disrupt
into its constituent.
b) if EB< 0 then the nucleus is not stable & it will disintegrate by itself.
MASS DEFECT (∆𝐦): The difference between the sum of the masses of constituent
Nucleon & mass of a nucleus is called the Mass defect.
The total mass of all the constituent Nucleon is always greater than the mass of the
nucleus.
Mass defect = sum of masses of the constituent nucleon - Rest mass of nucleus
∆𝐦 =(Zmp+Nmn) - M
∆𝐦 =[Zmp+(A-Z)mn]- M
where, Z = atomic number mp = mass of proton mn = mass of neutron
M= rest mass of nucleus
PACKING FRACTION (PF): The mass defect per nucleon is called Packing Fraction. It is the measure of comparative stability of an atom.
Packing fraction (PF) = 𝐦𝐚𝐬𝐬 𝐝𝐞𝐟𝐞𝐜𝐭(∆𝐦)/𝐦𝐚𝐬𝐬 𝐧𝐮𝐦𝐛𝐞𝐫(𝐀)
a) If PF > 0 then the nucleus is not stable.
b) If PF < 0 then nucleus is more stable
BINDING ENERGY or NUCLEAR BINDING ENERGY: The Binding energy of a nucleus is
defined as the minimum energy required to separate the nucleon from the nucleus far
apart to form stable nucleus. It is also defined as the minimum energy required to bind
the nucleons to form a stable nucleus. It is equivalent to mass defect. Thus mass defect
is a measure of the Binding energy of the nucleus.
∆𝒎 = (𝐙𝐦𝐩 + 𝐍𝐦𝐧) − 𝐌 ……………..(1)
If ∆𝐦 is in Kg then Binding energy of nucleus is given by
EB=(∆𝐦 ) c2
EB= [(𝐙𝐦𝐩 + 𝐍𝐦𝐧) − 𝐌 ] c2 ……………….. (2)
If ∆𝐦 is in amu or u , then Binding energy of nucleus is given by
EB=(∆𝐦 ) X 931 Mev ……………….. (3)
a) If EB> 0 then the nucleus is stable & external energy must be supplied to disrupt into its constituent.
b) If EB< 0 then the nucleus is not stable & it will disintegrate by itself
BINDING ENERGY PER NUCLEON & BINDING ENERGY CURVE:
a) BINDING ENERGY PER NUCLEON: It is also called average Binding Energy or Specific Binding energy. The Binding energy per Nucleon of a nucleus is called Binding energy per Nucleon. B.E=𝑬𝑩/𝑨=[(𝐙𝐦𝐩+𝐍𝐦𝐧) − 𝐌 ] 𝐜𝟐//𝐴………….(4) : If mass defect is in Kg
B.E=𝑬𝑩/𝑨=∆𝒎/𝑨x 931 MeV………….(5): if mass defect is in amu or u.
• It explains the stability of the nucleus
• The greater the value of Binding energy per nucleon , the more stable is the nucleus & vice versa
• It is the average Binding energy of a nucleus.
b) BINDING ENERGY CURVE: The graph between Binding energy per nucleon & mass number of different nuclei is known as Binding energy curve. The Binding energy curve is shown in fig 1.
• The curve reaches a peak of about 8.795MeV/nucleon at A=62, (Nickel). It is the measurement of stability of a nucleus
• B.E. per nucleon og light nuclei (1 H2) as well as heavy nuclei( 92U238) is small. It shows that they are relatively less stable.
• The nuclei in the middle range (about A=50 to 80) have highest B.E. per nucleon, which shows that they are most stable.
• The peak in the graph indicates that the nuclei 2He4, 6C12, 8O16 etc are relatively more stable than the other neighboring nuclei.
• 28Ni62 has largest B.E. per nucleon ( 8.795 MeV/A) of all nuclides , followed by 26Fe58 & 26Fe56 ( 8.792 MeV/A & 8.790 MeV/A respectively) & 1 H2 has lowest B.E. per nucleon (1.12MeV/A) of all nuclides.
• Nearly all stable nuclides from the lightest to the most massive B.E per nucleon in the range of 7 to 9 MeV/nucleon.
• If a heavy nucleus is split into two medium sized nuclei (Nuclear fission ), each of the new nuclei will have more B.E per nucleon than the original nucleus. for instance, if 92U238 is splitted , B.E per nucleon increases from 7.6 MeV from 8.4 MeV.
• If two light nuclei combines to give a single heavier nucleus (Nuclear fusion) ,then the newly formed nuclei have more B.E per nucleon than the original nucleus.
NUCLEAR FORCE:
➢ The nuclear force is a force that acts between the protons & neutrons of an atom.
➢ It is the force that binds the neutrons & protons in a nucleus together.
➢ The nuclear forces are the strongest force known to Physics.
➢ They are short range forces. They only act in the range of 10-15m.
➢ They are non conservative force (energy released is not recovered) & charge
independent forces.
➢ It is 100 times that of electrostatic force and 1038 times that of gravitational
force.
NUCLEAR FISSION:
The phenomenon in which a heavy nucleus splits up into two fragments of nearly comparable masses & a large amount of energy is released is called Nuclear Fission. The phenomenon was discovered by two German scientist s Fritz Stresemann & Otto Hahn in 1939.
When a Uranium nucleus 92U235 is bombarded by slow neutron 0n1 then it splits into 56Ba141 & 36Kr92 along with three neutrons & huge amount of energy is released.92U235 + 0n1 → 56Ba141 + 36Kr92 +30n1 + Q where Q is energy released in the process is about 200MeV.
Here, the total initial mass of 92U235 & 0n1 is greater than the total final mass of 56Ba141,36Kr92 & 30n1. The decrease in mass is converted into fission energy Q according to mass
- energy relation (E=mc2).
ENERGY RELEASED IN FISSION:
Mass of neutron, 0n1 = 1.008665 u
Mass of Uranium, 92U235 = 235.045733 u
Mass of Barium, 56Ba141= 140.917704 u
Mass of krypton, 36Kr92 = 91.8854 u
Mass of reactant = 1.008665 +235.045733 = 236.054398 u
Mass of product = 140.917704 + 91.8854 + 3x 1.008665 = 235.829095 u
Difference in mass(∆𝑚) = Mass of reactant - Mass of product
=236.054398 u - 235.829095 u =0.225303 u
Hence, energy released in a fission E= ∆𝑚 x 931 MeV = 0.225303 x 931 MeV= 209.76 MeV
≈ 200 MeV
Again B.E. per nucleon = 𝐸/𝐴=200𝑀𝑒𝑉/235= 0.8 MeV
NOTE: The energy released by 1kg of 92U235 is about 8.2 x 1013 J or 2.27 x 107 Kwh, which is equivalent to energy liberated in an explosion of 3000 tons of coal.
Now , one mole of 92U235 has mass 235 gram & contains 6.023 x 1023 atoms. i.e. 235 gm of 92U
235 contains 6.023 x 1023 atoms
so, 1 gm of 92U235 contains 6.023 x 1023/235atoms
∴1000 gm of 92U235 contains 6.023 x 1023/235 x 1000 atoms = N(say)
Hence , fusion of 1kg of 92U235,the energy releases is
E= Nx Q = 6.023 x 10^23/4x 1000 a x 24MeV =5.8x 1014J = 1.6 x 108 KWh
This is equal to energy released by burning million tons of coal.
Q-Value of a nuclear reaction: It indicates that the energy releases or absorbs during the nuclear reaction. It is based on Einstein mass energy relation (E=mc2). The energy change occurring in a nuclear reaction is termed as Q- value of the nuclear reaction.
✓ If Q-value > 0 then energy is released & reaction I called Exothermic or Exoergic reaction.
✓ If Q- value <0 then energy is absorbed & reaction is called Endothermic or Endoergic reaction.
NOTE: A neutron having energy less than 10 eV is called slow neutron.
[ 1KWh= 3.6X106J ]
MULTIPLICATION FACTOR (K): The ratio of number of secondary neutrons produced to the number of initial neutrons is called multiplication factor (K).
I.e. K= 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒏𝒆𝒖𝒕𝒓𝒐𝒏 𝒊𝒏 𝒂𝒏𝒚 𝒈𝒆𝒏𝒆𝒓𝒂𝒕𝒊𝒐𝒏 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒏𝒆𝒖𝒕𝒓𝒐𝒏 𝒊𝒏 𝒑𝒓𝒆𝒗𝒊𝒐𝒖𝒔 𝒈𝒆𝒏𝒆𝒓𝒂𝒕𝒊𝒐𝒏
If K>1, the chain reaction is dying down or subcritical.
IF K<1, the chain reaction is steady or critical.
If K=1, the chain reaction is building or supercritical.
CRITICAL SIZE & CRITICAL MASS: The minimum size of fissionable material to sustain chain reaction is called critical size. The mass corresponding to critical size is called critical mass. If the size of fissionable material is less than critical size then a chain reaction is not possible.
CHAIN REACTION: The fission reaction which continues until all the fissionable material is disintegrated is called chain reaction. It is self propagating reaction. e.g. When 92U235 is bombarded with slow neutron ( 0n1), three additional neutrons are released with emission of about 200MeV energy. These three neutrons strike three other Uranium nuclei, producing 9 neutrons & excess of energy. This process is happening till the whole fissionable material is disintegrated & huge amount of energy is released.
TYPES OF CHAIN REACTION:
a) UNCONTROLLED CHAIN REACTION: It is a chain reaction in which the energy produced cannot be controlled. In this reaction, the number of neutron goes on increasing geometrically & such a reaction proceeds very quickly with the releasing of huge amount of energy. So this reaction is also called explosive chain reaction. This principle is used in Atom Bomb.
b) CONTROLLED CHAIN REACTION: It is a chain reaction in which energy produced can be controlled in a desired level. In this reaction, the moderator [like graphite, heavy water (D2O) ] is used for slowing down the fast neutrons. This Principle is used in
Nuclear reaction.
DIFFERENCE BETWEEN NUCLEAR FISSION & NUCLEAR FUSION:
NUCLEAR FISSION:
a) Splitting of a heavy nucleus into light nuclei.
92U235 + 0n1 → 56Ba141 + 36Kr92 +30n1 + Q where Q is energy released
b) Energy released per fission of Uranium is around 200MeV
c) Energy released per nucleon is about 0.8MeV
d) Controlled fission reaction is the possible & is being used for constructive purpose in nuclear reactor.
e) Uncontrolled fission reaction is principle of atom bomb
f) No restriction of temperature
g) sources for nuclear fission are limited
h) A minimum mass known as critical mass is required to start fission
i) It is induced by neutrons
j) Fuel is either in solid or in liquid state
k) It is a single stage reaction & quick process
l) By products are harmful m) It produces nuclear pollution.
NUCLEAR FUSION:
a) Combination of lighter nuclei to form a heavy & stable nucleus
i.e. 1H2 + 1H2 →2He4+Q ,where Q is energy released
b) Energy released per fusion is around 24MeV
c) Energy released per nucleon is 6MeV
d) Controlled fusion reaction is not discovered yet.
e) Uncontrolled fusion reaction is the principle of Hydrogen Bomb
f) About 107K temperature is needed to start fusion
g) Sources for nuclear fusion reaction are almost unlimited.
h) No minimum mass is required for nuclear fusion.
i) It is Induced by protons
j) Fuel is in Plasma state
k) It is multi-stage reaction
l) By products are safe m) It produces thermal pollution.
DIFFERENCE BETWEEN CHEMICAL REACTION & NUCLEAR REACTION:
Chemical reaction:
I. In this reaction , only outer most electrons take part while nuclear composition remains the same
II. It may be reversible
III. No new atoms are formed
IV. Total number of atoms should be conserved
V. It is influenced by the change in pressure & temperature
Nuclear reaction:
I. In this reaction, nucleus of an atom is involved so nuclear composition is changed.
II. It is irreversible
III. New nuclei are formed
IV. Total mass number & atomic number are conserved
V. It is not influenced by the change in pressure & temperature
Dalton’s Model of an atom: John Dalton, a British school teacher, published his theory about atoms in 1808. His findings were based on experiments and the laws of chemical combination.
Postulates
1. All matter consists of indivisible particles called atoms.
2. Atoms of the same element are similar in shape and mass, but differ from the atoms of other elements.
3. Atoms cannot be created or destroyed.
4. Atoms of different elements may combine with each other in a fixed, simple,whole number ratios to form compound atoms.
5. Atoms of same element can combine in more than one ratio to form two or more compounds.
6. The atom is the smallest unit of matter that can take part in a chemical reaction.
Drawbacks of Dalton's Atomic Theory
• The indivisibility of an atom was proved wrong: an atom can be further subdivided into protons, neutrons and electrons. However an atom is the smallest particle that takes part in chemical reactions.
• According to Dalton, the atoms of same element are similar in all respects. However, atoms of some elements vary in their masses and densities. These atoms of different masses are called isotopes. For example, chlorine has two isotopes with mass numbers 35 and 37.
• Dalton also claimed that atoms of different elements are different in all respects. This has been proven wrong in certain cases: argon and calcium atoms each have an atomic mass of 40 amu. These atoms are known as isobars.
• According to Dalton, atoms of different elements combine in simple whole number ratios to form compounds. This is not observed in complex organic compounds like sugar (C12H22O11).
• The theory fails to explain the existence of allotropes; it does not account for differences in properties of charcoal, graphite, diamond.
Thomson’s Model of atom:
According to J.J. Thomson, the electron is a constituent of all matter. It has a negative
charge of 1.602X10-19C and a mass of 9.11X10-31Kg at rest. Electrons are distributed along
with positive charges in an atom in a sphere of radius 10-10m like a plum pudding. Atom is
electrically neutral as there are an equal number of positive and negative charges.
Failure of Thomson’s Atomic Model:
1) It could not explain the origin of spectral series as in case of a hydrogen atom.
2) It could not explain the largest angle of scattering of α-particles from the metal foil, as observed by Rutherford.
Rutherford’s Alpha Scattering Experiment:
Rutherford’s conducted an experiment by bombarding a thin sheet of gold with α-particles and then studied the trajectory of these particles after their interaction with the gold foil.
In this experiment, high energy streams of α-particles from a radioactive source(Polonium, R) kept inside a thick lead box. This collimated beam is then allowed to fall at a thin sheet (100 nm thickness) of gold F. While passing through the gold foil, the α-particles are scattered through different angles. The scattered α-particles are in a particular direction are allowed to strike on a screen coated with zinc sulphide. When α-particle is incident on zinc sulphide, it produces fluorescence and is detected with the help of Microscope M.
1. Most of the α-particles pass straight (without any deflection) through the gold foil, and hence most of the space in an atom is empty.
2. Some of the α-particles were deflected by very small angles, and hence the positive charge in an atom is not uniformly distributed. The positive charge in an atom is concentrated in a very small volume.
3. There were a few α-particles that were deflected by very large angles because α-particles were being repelled by a massive positive charge concentrated in a very small region of space.
4. Very few of the α-particles were deflected back, that is only a few α-particles had nearly 180° angle of deflection. So the volume occupied by the positively charged particles in an atom is very small as compared to the total volume of an atom.
Hence, this experiment showed that the atom is mostly empty with a tiny,dense, and positively-charged nucleus.
Rutherford Atomic Model
Based on the above observations and conclusions, Rutherford proposed the atomic structure of elements. According to the Rutherford atomic model:
1. The positively charged particles and most of the mass of an atom was concentrated at its centre in a very tiny region of the order 10-15m called as a nucleus.
2. Rutherford model proposed that the negatively charged electrons surround the nucleus of an atom. He also claimed that the electrons surrounding the nucleus revolve around it with very high speed in circular paths( orbits).
3. The total positive charge of nucleus is equal to the total negative charge on electron. Therefore an atom as a overall is neutral.
4. Electrons being negatively charged and nucleus being a densely concentrated mass of positively charged particles are held together by a strong electrostatic force of attraction.
Limitations of Rutherford Atomic Model:
1) It could not explain the stability of an atom.
2) It did not say anything about the arrangement of electrons in an atom.
3) It failed to explain the line spectra of Hydrogen.
1) What is Rutherford alpha particle scattering experiment?
Ans:This is also known as Rutherford's gold foil experiment or Geiger Marsden alpha scattering experiment. This is the experiment which led Rutherford to conclude that an atom is mostly empty and understand the concept of nucleus. Its outcome discarded the Thomson model of atom.
2) What were the conclusions of Rutherford's alpha particle scattering experiment?
Ans: Most of the space inside the atom is empty because most of the α-particles passed through the gold foil without getting deflected. Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space.
3) What were the results of the alpha particle scattering experiment?
Ans: Conclusions and arguments: The results of this experiment were not in sync with the plum-pudding model of the atom as suggested by Thomson. Rutherford concluded that since alpha particles are positively charged, for them to be deflected back, they needed a large repelling force.
4) Why did Rutherford use alpha particles in his experiment?
And:Rutherford wanted to check the postulates of Thompson and hence he used POSTIVILEY CHARGED ALPHA PARTICLES IN HIS SCATTERING EXPERIMENT. ... almost all of the radiations passed through the gold foil without any deviation and only a few, maybe 1 in 10,000 , suffered a scattering, which indicates that atoms are hollow.
5) Why did Rutherford use gold foil?
Ans: Rutherford lacked the endurance for this work, which is why he left it to his younger colleagues. For the metal foil, they tested a variety of metals, but they preferred gold because they could make the foil very thin, as gold is very malleable.
6) Why Alpha particles are deflected?
Ans:Most of the alpha particles did pass straight through the foil. The atom being mostly empty space. A small number of alpha particles were deflected by large angles (> 4°) as they passed through the foil. ... Like charges repel, so the positive alpha particles were being repelled by positive charges.
7) How did Rutherford get alpha particles?
Ans:The origin of alpha particles is the nucleus of an atom. Double-ionizing a helium atom does not produce an alpha particle. As pointed out, the source of alpha particles that was used by Rutherford and his associates was the radioactive decay of Radium, where alpha particles are produced in abundance.
8) What is the most important discovery made by the alpha particle scattering experiment? of Rutherford's alpha particle scattering experiment?
Ans:Rutherford's alpha-particle scattering experiment was responsible for the discovery of atomic nucleus. Nucleus contains most of the mass. Most of the atom has empty.
9) What did Rutherford's gold foil experiment prove?
Ans: Thomson's plum pudding model of the atom had negatively-charged electrons embedded within a positively-charged "soup." Rutherford's gold foil experiment showed that the atom is mostly empty space with a tiny, dense, positively-charged nucleus. Based on these results, Rutherford proposed the nuclear model of the atom.
10) What was the thickness of gold foil used in Rutherford's experiment?
Ans:Because it is unusually ductile, gold can be made into a foil that is only 0.00004cm thick. When this foil was bombarded with -particles, Geiger found that the scattering was small, on the order of one degree. These results were consistent with Rutherford's expectations
11) What were two conclusions of the gold foil experiment?
Ans:Conclusion of Rutherford's scattering experiment: Most of the space inside the atom is empty because most of the α-particles passed through the gold foil without getting deflected. Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space
THE END