MCQ
The frequency of radiation emitted when the electron falls from $n = 4$ to $n = 1$ in a hydrogen atom will be (Given ionization energy of $H$ $ = 2.18 \times {10^{ - 18}}J\,\,\,{\rm{ato}}{{\rm{m}}^{ - 1}}$ and $h = 6.625 \times {10^{ - 34}}\,Js)$
  • $3.08 \times {10^{15}}{s^{ - 1}}$
  • B
    $2.00 \times {10^{15}}{s^{ - 1}}$
  • C
    $1.54 \times {10^{15}}{s^{ - 1}}$
  • D
    $1.03 \times {10^{15}}{s^{ - 1}}$

Answer

Correct option: A.
$3.08 \times {10^{15}}{s^{ - 1}}$
a
(a) ${E_{{\rm{ionisation}}}} = {E_\infty } - {E_n} = \frac{{13.6Z_{eff}^2}}{{{n^2}}}\,eV$

= $\left[ {\frac{{13.6{Z^2}}}{{n_2^2}} - \frac{{13.6{Z^2}}}{{n_1^2}}} \right]$

$E = h\nu = \frac{{13.6 \times {1^2}}}{{{{(1)}^2}}} - \frac{{13.6 \times {1^2}}}{{{{(4)}^2}}}$; $h\nu = 13.6 - 0.85$

$h = 6.625 \times {10^{ - 34}}$

$\nu = \frac{{13.6 - 0.85}}{{6.625 \times {{10}^{ - 34}}}} \times 1.6 \times {10^{ - 19}}$ = $3.08 \times {10^{15}}{s^{ - 1}}$.

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