Question

Ionization energy of a hydrogen-like ion A is greater than that of another hydrogen-like ion B. Let, r, u, E and L represent the radius of the orbit, speed of the electron, energy of the atom and orbital angular momentum of the electron respectively. In ground state
A. \[{r_A} > {r_B}\]
B. \[{u_A} > {u_B}\]
C. \[{L_A} > {L_B}\]
D. None of these.

Answer 1

Hint: The energy required to remove an electron from any state is equal to the total energy of the electron in the orbit. The total energy of the electron is due to the kinetic energy and the electrostatic potential energy.

Formula used:
\[{E_n} = - 13.6\left( {\dfrac{{{Z^2}}}{{{n^2}}}} \right)eV\]
where \[{E_n}\] is the total energy of the electron in a hydrogen-like atom of atomic number Z in the quantum state number n.

Complete step by step solution:
It is given that the ionization energy of A is greater than that of B. As per the Bohr model of atom, the ionization energy of the electron from hydrogen like atom is,
\[{E_i} = \left( {13.6eV} \right)\left( {\dfrac{{{Z^2}}}{{{n^2}}}} \right)\]
If the atomic numbers of A and B are \[{Z_A}\] and \[{Z_B}\] respectively. The electrons are in ground state, i.e. \[{n_A} = {n_B} = 1\] then the ionization energy of the atoms will be,
\[{E_{iA}} = \left( {13.6eV} \right)\left( {\dfrac{{Z_A^2}}{{n_A^2}}} \right) \\ \]
\[\Rightarrow {E_{iA}} = \left( {13.6eV} \right)\left( {\dfrac{{Z_A^2}}{1}} \right) \\ \]
\[\Rightarrow {E_{iA}} = \left( {13.6eV} \right)\left( {Z_A^2} \right) \\ \]
\[\Rightarrow {E_{iB}} = \left( {13.6eV} \right)\left( {\dfrac{{Z_B^2}}{{n_B^2}}} \right) \\ \]
\[\Rightarrow {E_{iB}} = \left( {13.6eV} \right)\left( {Z_B^2} \right)\]
It is given that the ionization energy of A is greater than that of B. So, the atomic number of A is greater than that of B.

The expression for the speed of electron is,
\[v = \sqrt {\dfrac{{Z{K_e}{e^2}}}{{{m_e}{r_n}}}} \]
So, the relation between the speed of the electron and the atomic number of atom is proportional to the square root of atomic number,
\[u \propto \sqrt Z \]
As the atomic number of A is greater than that of B, so the speed of electron in atom A is greater than that of B.
\[{u_A} > {u_B}\]
The expression for the radius of nth orbit is,
\[{r_n} = \dfrac{{{n^2}}}{Z}\left( {\dfrac{{{h^2}}}{{{K_e}{e^2}{m_e}}}} \right) \\ \]
\[\Rightarrow r \propto \dfrac{1}{Z}\]

As the radius of orbit is inversely proportional to the atomic number. So the radius of orbit for atom B is greater than that of atom A.
\[{r_A} < {r_B}\]
The angular momentum of an electron in nth orbit is given as,
\[L = \dfrac{{nh}}{{2\pi }}\]
As the electrons in both the atoms are in the same orbit, so the angular momentum is equal.
\[{L_A} = {L_B}\]

Therefore, the correct option is B.

Note: The energy required to remove a loosely bound electron from the influence of the nucleus of the atom. Closest the electron is from the nucleus of the atom, greater will be the required energy to remove the electron from the influence of the nucleus of the atom.