MCQ
The ratio of difference between $1^{st}$ and $2^{nd}$ Bohr orbits energy to that between  $2^{nd}$ and $3^{rd}$ orbits energy is
  • A
    $0.5$
  • B
    $\frac{1}{3}$
  • $5.4$
  • D
    $\frac{5}{27}$

Answer

Correct option: C.
$5.4$
c
For H-atom

$ E_{3} =-\frac{13.6}{(3)^{2}}=-{13.6}{9} $

$=-1.5 \mathrm{eV} $

$ E_{2} =-\frac{13.6}{(2)^{2}}=-{13.6}{4} $

$=-3.4 \mathrm{eV} $

$ E_{1} =-{13.6}{(1)^{2}}=-{13.6}{1} $

$=-13.6 \mathrm{eV} $

$Now E_{2}-E_{1}=(-3.4)-(-13.6) $

$=13.6-3.4=10.2 \mathrm{eV} $

$ E_{3}-E_{2} =(-1.5)-(3.4) $

$=3.4-1.5=1.9 \mathrm{eV} $

$ \therefore \quad \frac{E_{2}-E_{1}}{E_{3}-E_{2}} =\frac{10.2}{1.9}=5.36 $

$=5.4=\frac{27}{5} $

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