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

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 12 - 12. atoms4 Marks
MCQ
An electron jumps from $5^{th}$ orbit to $4^{th}$ orbit of hydrogen atom. Taking the Rydberg constant as ${10^7}$ per metre. What will be the frequency of radiation emitted
  • A
    $6.75 \times {10^{12}}\,Hz$
  • B
    $6.75 \times {10^{14}}\,Hz$
  • ✓
    $6.75 \times {10^{13}}\,Hz$
  • D
    None of these
✓

Answer

Correct option: C.
$6.75 \times {10^{13}}\,Hz$
c
(c) By using $\nu = RC\,\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]$

$ \Rightarrow \nu = {10^7} \times (3 \times {10^8})\,\left[ {\frac{1}{{{4^2}}} - \frac{1}{{{5^2}}}} \right]$= $6.75 ×10^{13} Hz$

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