```ATOMIC TERM
Nuclide:
Various species of atoms in general.
Nucleons:
Sub-atomic particles in the nucleus of an atom, i.e., protons and neutrons.
Isotopes:
Atoms of an element with the same atomic number but different mass number.
Mass number (A):
Sum of the number of protons and neutrons, i.e., the total number of nucleons
Atomic number (Z):
The number of protons in the nucleus of an atom. This, when subtracted from A, gives the number of neutrons.
Isobars:
Atoms, having the same mass numbers but different atomic numbers, e.g., 15P32 and16S32.
Isotones:
Atoms having the same number of neutrons but different number of protons or mass number, e.g., 146C, 157N.
Isoelectronic species:
Atoms molecules or ions having the same number of electrons, e.g., N2, CO, CN-.
Nuclear isomers:
Atoms with the same atomic and mass numbers but different radioactive properties, e.g., uranium X (half life 1.4 min) and uranium Z (half life 6.7 hours).
Atomic mass unit:
Exactly equal to 1/12th of the mass of 6C12 atom.
(a.m.u.): 1 a.m.u. = 1.66 x10–24 g ≈ 931.5 MeV
Illustration:
The ion that is isoelectronic with CO is
(A)    CN-                           (B)    O2+
(C)    O2-                            (D)    N2+
Solution:
Both CO and CN- have 14 electrons. Hence (A) is correct

Illustration:-
Find the number of neutrons in a neutral atom having atomic mass 23 and number of electrons 11.
Solution:
Number of protons = number of electrons = 11
Number of neutrons = atomic mass – number of protons =23–11 = 12

Illustration
How many protons, electrons and neutrons are present in 0.18 g1530?
Solution:
No. of neutrons in one atom = (30 – 15) = 15
0.18 g 3015P = 0.18 / 30  = 0.006 mole
Now, number of  3015P atoms in 0.006 mole = 0.006 x 6.02 x 1023
Number of electrons in 0.006 mole of 3015P = Number of protons in 0.006 mole 3015P = 15 x 0.006 x 6.02 x 1023 = 5.418 x 1022 and number of neutrons = 5.418 x 1022
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