If ${\lambda _{\max }}$ is $6563\,\mathring{A}$ for the Balmer series of a particular atom,then the wavelength of the second line for the Balmer series will be:

  • A
    $\lambda = \frac{16}{3R}$
  • B
    $\lambda = \frac{36}{5R}$
  • C
    $\lambda = \frac{4}{3R}$
  • D
    None of the above

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