Explain isotopes, isobar, isotones and isomers by giving examples.
Explain isotopes, isobar, isotones and isomers by giving examples.
Isotopes : The atoms which have atomic number $Z$ is same but atomic mass number $(A)$ is different then such type of atoms are called the isotopes of each other.
Isotopes of hydrogen are ${ }_{1} \mathrm{H}^{1},{ }_{1} \mathrm{H}^{2},{ }_{1} \mathrm{H}^{3}$ in ${ }_{1} \mathrm{H}^{1}$ there is one proton but having none neutron. In ${ }_{1} \mathrm{H}^{2}$ there is one proton and one neutron and in ${ }_{1} \mathrm{H}^{3}$ there is one proton and two neutrons. Gold has 32 isotopes whose mass numbers range from $\mathrm{A}=173$ to $\mathrm{A}=204$.
As the atoms of isotopes have identical electronic structure they have identical chemical behaviour and are placed in the same location in the periodic table.
In addition ${ }_{6}^{12} \mathrm{C},{ }_{6}^{13} \mathrm{C},{ }_{6}^{14} \mathrm{C}$ are the isobaric nuclide of carbon and ${ }_{92}^{233} \mathrm{U},{ }_{92}^{235} \mathrm{U},{ }_{92}^{238} \mathrm{U}$ are the isobaric nuclide of uranium and ${ }_{3}^{3} \mathrm{Li}$ and ${ }_{3}^{4} \mathrm{Li}$ are the isotopes of lithium.
Similar Questions
The radius of $R$ of a nucleus of mass number $A$ can be estimated by the formula $R =\left(1.3 \times 10^{-15}\right) A ^{1 / 3}\, m .$ It follows that the mass density of a nucleus is of the order of
$\left( M _{\text {prot. }} \cong M _{\text {neut. }}=1.67 \times 10^{-27} kg \right)$
The radius of $R$ of a nucleus of mass number $A$ can be estimated by the formula $R =\left(1.3 \times 10^{-15}\right) A ^{1 / 3}\, m .$ It follows that the mass density of a nucleus is of the order of
$\left( M _{\text {prot. }} \cong M _{\text {neut. }}=1.67 \times 10^{-27} kg \right)$
- [JEE MAIN 2020]
$STATEMENT-1$: The plot of atomic number ( $\mathrm{y}$-axis) versus number of neutrons ( $\mathrm{x}$-axis) for stable nuclei shows a curvature towards $x$-axis from the line of $45^{\circ}$ slope as the atomic number is increased. and $STATEMENT-2$: Proton-proton electrostatic repulsions begin to overcome attractive forces involving protons and neutrons and neutrons in heavier nuclides.
$STATEMENT-1$: The plot of atomic number ( $\mathrm{y}$-axis) versus number of neutrons ( $\mathrm{x}$-axis) for stable nuclei shows a curvature towards $x$-axis from the line of $45^{\circ}$ slope as the atomic number is increased. and $STATEMENT-2$: Proton-proton electrostatic repulsions begin to overcome attractive forces involving protons and neutrons and neutrons in heavier nuclides.
- [IIT 2008]
Highly energetic electrons are bombarded on a target of an element containing $30$ neutrons. The ratio of radii of nucleus to that of Helium nucleus is ${14^{1/3}}$. The atomic number of nucleus will be
Highly energetic electrons are bombarded on a target of an element containing $30$ neutrons. The ratio of radii of nucleus to that of Helium nucleus is ${14^{1/3}}$. The atomic number of nucleus will be
A heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio $8 : 1$. The ratio of radii of the fragments is
A heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio $8 : 1$. The ratio of radii of the fragments is
Why do stable nuclei never have more protons than neutrons ?
Why do stable nuclei never have more protons than neutrons ?