NEET-XII-Physics

43: Bohr's Model and Physics of the Atom

with Solutions - page 5
Qstn# iv-11 Prvs-QstnNext-Qstn
  • #11
    Find the maximum Coulomb force that can act on the electron due to the nucleus in a hydrogen atom.
    Ans : Charge on the electron, q1 = 1.6`` \times ``10 `` -19`` C
    Charge on the nucleus, q2 = 1.6`` \times ``10 `` -19`` C
    Let r be the distance between the nucleus and the electron.
    Coulomb force `` \left(F\right)`` is given by
    `` F=\frac{{q}_{1}{q}_{2}}{4\,\mathrm{\,\pi \,}{\in }_{0}{r}^{2}}`` .....(1)
    Here, q1 = q2 = q = 1.6`` \times ``10 `` -19`` C
    Smallest distance between the nucleus and the first orbit, r = 0.53`` \times ``10`` -``10 m
    `` K=\frac{1}{4{\,\mathrm{\,\pi \epsilon \,}}_{0}}`` = 9`` \times ``109 Nm2C`` -``2
    Substituting the respective values in (1), we get
    `` F=\frac{\left(9\times {10}^{9}\right)\times \left(1.6\times {10}^{-19}\right)\times \left(1.6\times {10}^{-19}\right)}{{\left(0.53\times {10}^{-10}\right)}^{2}}``
    `` =\frac{1.6\times 1.6\times 9\times {10}^{-9}}{{\left(0.53\right)}^{2}}=82.02\times {10}^{-9}``
    `` =8.2\times {10}^{-8}\,\mathrm{\,N\,}``
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