Though there are several elementary particles, only relevant particles in earthly life and in nuclear reactors, are photons and particles which constitute material, i.e. protons, neutrons, and electrons. Among these, proton and neutron have about the same mass. Though, mass of electron is only 0.05% that of the two particles. Proton has positive charge and its absolute value is same as electric charge of one electron (elementary electric charge). Proton and neutron are known as nucleons and they comprise nucleus. The atom is comprised of nucleus and electrons which circle nucleus because of Coulomb attraction.
Species of atoms and nuclei are known as elements and nuclides respectively. Element is determined by its proton number (number of protons). Proton number is usually known as atomic number and is signified by Z. A nuclide is determined by both proton number and neutron number, number of neutrons denoted by N. Sum of proton number and neutron number, that is, nucleon number, is known as mass number and is signified by A (A=Z+N).
To identify nuclide, A and Z are generally added on left side of atomic symbol as superscript and subscript respectively. For instance, there are two representative nuclides for uranium, explained as 23592U and 23892U. If atomic symbol is given, atomic number can be exclusively determined; therefore Z is frequently omitted like 235U and 238U. Chemical properties of atom are determined by atomic number, so even if mass numbers of nuclei are different, if atomic numbers are same, their chemical properties are similar. These nuclides are known as isotopic elements or isotopes. If mass numbers are same and the atomic numbers are different, they are known as isobars. If neutron numbers are same, they are known as isotones.
The nuclide is any nuclear specie with combination of neutrons and protons i.e. AXZ , A= atomic mass, Z = atomic number = number of proton = number of electrons while number of neutron, N = (A - Z).
P(r) = Ρ0/1+e(r-R)/b
Where Ρ0 = density at nuclear centre
R= radius at which Ρ falls to Ρ0/2
b= estimates how rapidly density falls to zero at nuclear surface.
R = radius of nucleus
R = r0A1/3
R = radius of nucleus and r0 = 1.23x10-15m
This signifies that:
1 a.m.u. = 1.6660053 x 10-27kg = 931MeV
To show that electron is not constituent of nucleus
Uncertainty principle is applied here.
Δp ≈ (h/2π)/Δx = 1.255 x 10-34/2x10-14 = 5.275 x 10-21kgms-1
If this is uncertainty in momentum of electron, momentum itself should be at least comparable in magnitude.
The K.E. of electron of mass, m may be expressed as follows:
T = Ρ2/2m where m = 9 x 10-31kg
= 9 x 10-31kg x 1.6 x 10-19ev
1.44 x 10-49eV = (5.275 x 10-21)2/2x1.44x10-49 = 97MeV
It follows that if electrons are present inside nucleus, their K.E should be of the order 97MeV. But experimental data reveal that no electron in atom has energy greater than 4MeV.
Excess mass and packing fraction:
Excess mass is stated as difference between masses of nucleons (M) and atomic mass (A). This signifies that excess mass = (M - A), that can either be either positive or negative.
Packing fraction (f) can be stated as ratio of excess mass to atomic mass. This signifies that packing fraction is stated as (M -A)/A= f.
Alpha (α) scattering experiment led to discovery of nucleus of atom. Mass of atom appears to be concentrated at nucleus and it is surrounded by cloud of electrons that makes entire atom electrically neutral.
One of the goals of Rutherford's α-scattering is the determination of radius (R) of nucleus that is as α -particle approaches gold nucleus it slows down because of Coulomb force but later speeds up on its way-out.
Coulomb repulsive force in region close to scattering gold nucleus is provided by:
F = 2eZe/4πε0b2
Time of operation of force
Δt = b/v
Force produces momentum, Δp, that is perpendicular to direction of α-particle.
Δp = f.Δt
Δp = (2eZe/4πε0b2)(b/v)
θ = Δp/p = 2Ze2/4πε0bmv2
MAke b = R and θ = 1
R = (1/4πε0)(2Ze2/mv2)
R ≈ 10-14m Radius of nucleus, R is smaller than radius of atom.
Nuclear Binding Energy and Separation Energy:
Binding energy is energy which should be supplied to dissociate nucleus in separate nucleus or energy released when separated nucleons were assembled into nucleus.
B (A, Z) = [ZMH + NMN - M(A,Z)] 931MeV
Also, difference between actual nuclear mass and mass of whole individual nucleus is known as mass defect (Md) that is equal to W - M.
Binding energy is measure of cohesiveness of the nucleus which is between proton and neutron. Also, more helpful measure of cohesiveness is binding energy per nucleon.
Binding energy can also be written in terms of mass number or atomic mass number
B = AMN - Z(MN - MH) - M(A,Z)a.m.u
Dividing through by A
B = MN - Z/A(MN - MH) - M/A
Work essential to separate proton, neutron, deuteron or α -particle from nucleus is known as Separation Energy.
Sn = M[A - 1, Z] + Mn - M(A,Z)931MeV
Sp = M[A - 1, Z - 1] + MH - M(A,Z)931MeV
Previously, we considered the curve of binding energy per nucleon against mass number (B/A against A curve). The value of B/A is approximately constant and it is about 8 MeV/nucleon. This is about a million time higher than the binding energy of an electron in the hydrogen atom (which is 13.6 eV). In other words, the force that keeps the nucleus together is much stronger than the electrical force which keeps the atom together. Also, this nuclear force must be stronger than the electrical force between the protons since protons are bound in the nuclei. The nuclear force is often called the strong interaction, because it is the strongest of the four basic forces or interactions found in nature.
Magnetic Dipole and Electric Quadrupole Moments of Deuteron:
The deuteron is in an l = 0 state and so we don't expect any orbital angular momentum. One expects total magnetic moment of deuteron to be sum of magnetic moments because of spin motions of neutron and proton. Therefore, we expect value:
μ(neutron) + μ(proton) = 0.8797[(eh/2π)/2Mp] for deuteron magnetic moment.
Though, experimentally obtained value of magnetic moment of deuteron was:
Deuteron magnetic moment, μd = 0.8574[(eh/2π)/2Mp]
The difference is approx 2.5%, which is small but important. The deuteron is in l = 0 state is not completely correct. There should be some orbital momentum in deuteron nuclear system. This signifies that wave functions describing deuteron are not spherically symmetric.
Spherically symmetric wave functions have only r dependence and no (θ,Φ) dependence. This is what occurs in l = 0 state; wave function is function f r only and so, is spherically symmetric.
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