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

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 12. atoms4 Marks
MCQ
The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, $\lambda _1/\lambda _2$, of the photons emitted in this process is
  • ✓
    $20/7$
  • B
    $7/5$
  • C
    $9/7$
  • D
    $27/5$
✓

Answer

Correct option: A.
$20/7$
a
$\frac{1}{\lambda } = R\left( {\frac{1}{{n_f^2}} - \frac{1}{{n_i^2}}} \right);$ $\frac{1}{{{\lambda _1}}} = R\left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right)$

$\frac{1}{{{\lambda _1}}} = R\left( {\frac{7}{{9 \times 16}}} \right);$ $\frac{1}{{{\lambda _2}}} = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right)$

$=R\left(\frac{5}{4 \times 9}\right)$

$\frac{\lambda_{1}}{\lambda_{2}}=\frac{\frac{5}{36}}{\frac{7}{9 \times 16}}=\frac{20}{7}$

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