If the shortest wavelength of the hydrogen atom in the Lyman series is $x$, then the longest wavelength in the Balmer series of $He^+$ is:

  • A
    $\frac{9x}{5}$
  • B
    $\frac{36x}{5}$
  • C
    $\frac{x}{4}$
  • D
    $\frac{5x}{9}$

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