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प्रश्न
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
पर्याय
rA > rB
uA > uB
EA > EB
LA > LB
उत्तर
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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संबंधित प्रश्न
Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, a thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10−10 m).
(a) Construct a quantity with the dimensions of length from the fundamental constants e, me, and c. Determine its numerical value.
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