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) E_A > E_B
(d) L_A > L_B

Answer 1

(b) u_A > u_B
The ionisation energy of a hydrogen like ion of atomic number Z is given by
$V=(13.6\mathrm{eV})\times Z^2$

Thus, the atomic number of ion A is greater than that of B (Z_A > Z_B).
The radius of the orbit is inversely proportional to the atomic number of the ion.
∴ r_A > r_B
Thus, (a) is incorrect.
The speed of electron is directly proportional to the atomic number.
Therefore, the speed of the electron in the orbit of A will be more than that in B.
Thus, u_A > u_B is correct.
The total energy of the atom is given by
$E=-\frac{m{Z}^2{e}^2}{8{\in }_0{h}^2{n}^2}$
As the energy is directly proportional to Z², the energy of A will be less than that of B, i.e. E_A < E_B.
The orbital angular momentum of the electron is independent of the atomic number.
Therefore, the relation L_A > L_B is invalid.