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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
Options
rA > rB
uA > uB
EA > EB
LA > LB
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Solution
uA > uB
The ionisation energy of a hydrogen like ion of atomic number Z is given by
`V = (13.6 eV)xx Z^2`
Thus, the atomic number of ion A is greater than that of B (ZA > ZB).
The radius of the orbit is inversely proportional to the atomic number of the ion.
∴ rA > rB
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, uA > uB is correct.
The total energy of the atom is given by
`E=-(mZ^2e^2)/(8∈_0h^2n^2)`
As the energy is directly proportional to Z2, the energy of A will be less than that of B, i.e. EA < EB.
The orbital angular momentum of the electron is independent of the atomic number.
Therefore, the relation LA > LB is invalid.
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