If we express the energy of a photon in $KeV$ and the wavelength in $\mathring{A}$,then the energy of a photon can be calculated from the relation:

  • A
    $E = 12.4\,h\nu$
  • B
    $E = 12.4\,h/\lambda$
  • C
    $E = 12.4/\lambda$
  • D
    $E = h\nu$

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