An electron jumps from the $4th$ orbit to the $2nd$ orbit of a hydrogen atom. Given the Rydberg constant $R = 10^5 \text{ cm}^{-1}$. The frequency in $\text{Hz}$ of the emitted radiation will be:

  • A
    $\frac{3}{16} \times 10^5$
  • B
    $\frac{3}{16} \times 10^{15}$
  • C
    $\frac{9}{16} \times 10^{15}$
  • D
    $\frac{3}{4} \times 10^{15}$

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