HomeLearn Fast select Course
Show Unit/ChapterChange UnitChange SubjectLast Menu
NEET-XII-Physics

43: Bohr's Model and Physics of the Atom

with Solutions - page 3
Qstn# iii-5 Prvs-QstnNext-Qstn
  • #5
    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) rA > rB
    (b) uA > uB
    (c) EA > EB
    (d) LA > LB
    digAnsr:   b,a
    Ans : (b) uA > uB
    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 (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=-\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. EA < EB.
    The orbital angular momentum of the electron is independent of the atomic number.
    Therefore, the relation LA > LB is invalid.
    Page No 384: