Nucleus And It's Composition

NUCLEUS AND ITS COMPOSITION Nucleus and Nuclear Particles: Nucleus is small and positively charged part of an atom at the center where entire mass of the atom is concentrated. E. Rutherford and his co-workers conducted a series of scattering experiments, on the basis of which the existence of he nucleus was first proposed in1911. These experiments helped to understand the arrangement of sub atomic particles (electrons, protons and neutrons) in an atom. From the experimental observations it was concluded that the protons and neutrons are present in the nucleus where as electrons are present in the empty space around the nucleus. ATOMIC NUMBER: It is the numbers of electrons present out side the nucleus or the number of protons present in side the nucleus. It is denoted by (Z). Atomic Number (Z) = Number of electrons (e) OR Number of Protons (p) NUCLEONS: The sub atomic particles (protons and neutrons) of nucleus are collectively called nucleons. The total number of nucleons is denoted by A and is called mass number of the nucleus. Mass Number (A) = Number of Protons (p) + Number of Neutrons (n) REPRESENTATION OF NUCLIDE: The nucleus of any atom is represented by specifying the atomic number as a subscript at the left hand bottom of the atomic symbol and mass number as a superscript at the left hand top of the symbol. For example, carbon atom, its symbol is C; atomic number is 6, mass number is12. It is represented as 12c .Such symbols are called as nuclides. 6

NUCLEAR DIMENSIONS: • The radius of the nucleus is of the order of 10-12 and that of atom is 10-8 • The radii of the various nuclei can be calculated by the following relation r = R0 A1/3 Where; r= radius of nucleus A= mass number R0 = constant (1.4 x 10-24cm) •

Nuclear Dimensions are expressed in Fermi units, 1Fermi = 10-13cm.

•

Area of cross section of nucleus is measured in unit called barn (1barn = 10-24cm2) 5 • Radius of nucleus is 10 times smaller than that of atom. • Density is of the order of 1014 gcm-3. • Volume is of the order of 10-38 cm3. ISOTOPES, ISOBARS, ISOTONES, ISODIAPHERS AND NUCLEAR ISOMERS Isotopes: These are the atoms having same atomic number but different mass numbers. They contain same number of protons but different number of nucleons. For example: 11H ,12H ,13H are the isotopes of hydrogen 84

215

Po ,

84

211

Po are the isotopes of polonium.

Isobars: The atoms having different atomic numbers but same mass number are called isobars. They contain same number of nucleons but different number of protons. For example: 2040Ca , 1840Ar or 82211Pb , 82211Bi Calcium and Argon are isobars of each other having same mass number but different atomic number. Similarly Lead and Bismuth are isobars of each other. Isotones: These are the atoms having same number of neutrons but different number of nucleons. But by appearance they have different number atomic number and different mass number but the number of neutrons is same. For example: 1430Si , 1531P , 1632S all these nuclides have different mass number and atomic number. But if we calculate the number of neutron in each case it is coming out to bethe16. Because Number of neutrons = mass number (A) – atomic number (Z) Isodiaphers: These are the species which have different atomic number and different mass number but same isotopic number i.e. (A-2Z) or (n-Z) For example, 19 39 9 F and 19 K are isodiaphers because isotopic number is 1.

Nuclear Isomers: Those nuclear species which have same atomic number, same mass number but have different radioactive properties. The difference in the properties is due to the difference in their energy states. BINDING ENERGY OF NUCLEUS As we know, protons and neutrons all together are called nucleons. The sum of the individual masses of various particles in the nucleus must be equal to the nuclear mass. But actually it is not like so. There is a difference in the masses. So this difference in the actual nuclear mass and the expected nuclear mass (sum of the individual masses of nuclear particles) is called mass defect. According to Einstein the mass defect can be converted into equivalent energy with the help of Einstein equation (E = mc2). This energy that is equivalent to mass defect is responsible for holding the nucleon together is called Binding Energy of the nucleus. For example Calculation for the binding energy of helium nucleus: Helium nucleus has 2 protons and 2 neutrons. Mass of two free neutrons (2x1.00867u) = 2.01734 u Mass of two free protons (2x1.00728u) = 2.01456 u Sum of the masses of 2 free neutrons And 2 free protons = -----------------4.03190 u Observed masses of 2 protons and 2 neutrons present in a helium nucleus = 4.00150 u ----------------------Mass defect (Δm)

= 0.03040 u

Mass defect in gram = 0.03040 / 6.02 x 1023 gram (1u= 1/ 6.02 x 1023) ΔE (binding energy) = mc2 = 0.03040 x 10-3 / 6.023 x 1023 kg = 0.03040 x 10-3 x (2.998 x 108 ms-1) 2 / 6.023 x 1023

= 4.539 x 10-12 J = 4.539 x10-12 / 1.602 x 10-19 eV = 2.83 x 107 eV or 2.83 x 107 x 10-6 MeV = 28.3MeV (million electron volt) This is the binding energy of the helium nucleus. As helium has 4 nuclear particle, therefore binding energy per nucleon in He – 4 nucleus is 28.314 = 7.07 MeV. The binding energy per nucleon is a measure of stability of nucleus.

Direct method of calculating binding energy in MeV and in joules

Mass defect (Δm) in u or amu can be directly calculated by using formula

BINDING ENERGY ANDNUCLEAR STABILITY Binding energy helps in finding the thermodynamic stability of the nuclide. Larger the value of binding energy, larger will be the stability.

Variation of binding energy per nucleon with mass number

Study of the graph reveals the following points. 1) For lighter nuclides binding energy is less. 2) It increases sharply in the beginning and attains a maximum value of 8.5 MeV around mass number56. 3) In the central part it is reasonably constant then start decreasing gradually after mass number 100. 4) Binding energy for 4He, stability.

12

C, and

16

O is exceptionally high showing maximum

5) In case of heavier nuclides, the low value of binding energy shows the unstable nature so they undergo nuclear fission reaction to form nuclei of medium mass number and are relatively more stable. 6) The lighter nuclei may tend to undergo fusion to give nuclei of medium mass number

---------------------------------------------------------------