In a sequence of two consecutive steps, a hydrogen atom in its ground state absorbs two photons and finally reaches its $n=7$ excited state. The energy of the first incident photon is found to be 12.47 eV , with an error margin of $\pm 0.32 \mathrm{eV}$. The energy, in eV , of second absorbed photon is closest to

• Correct answer
  A

  $0.57$
• B

  $1.45$
• C

  $0.81$
• D

  $3.40$

Hence, the correct answer is option: 'A'.