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STD 12 - 12. atoms — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 12. atoms4 Marks
MCQ
A double charged lithium atom is equivalent to hydrogen whose atomic number is $3$. The wavelength of required radiation for emitting electron from first to third Bohr orbit in $L{i^{ + + }}$ will be.......$\mathop A\limits^o $ (Ionisation energy of hydrogen atom is $13.6\,eV$)
  • A
    $182.51$
  • B
    $177.17 $
  • C
    $142.25$
  • ✓
    $113.74$
✓

Answer

Correct option: D.
$113.74$
d
(d) ${E_n} = - 13.6\frac{{{Z^2}}}{{{n^2}}}eV.$ Required energy for said transition

$\Delta E = {E_3} - {E_1} = 13.6\;{Z^2}\left[ {\frac{1}{{{1^2}}} - \frac{1}{{{3^2}}}} \right]$

$ \Rightarrow \Delta E = 13.6 \times {3^2}\left[ {\frac{8}{9}} \right] = 108.8\;eV$

$ \Rightarrow \Delta E = 108.8 \times 1.6 \times {10^{ - 19}}J$

Now $\Delta E = \frac{{hc}}{\lambda } = 108.8 \times 1.6 \times {10^{ - 19}}$

$ \Rightarrow \lambda = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{108.8 \times 1.6 \times {{10}^{ - 19}}}} = 0.11374 \times {10^{ - 7}}m$

$ = 113.74{Å}$

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