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MMLU-Pro / 4331 / You are given that the ionization energy of a hydrogen atom is 13.527eV,…

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question

You are given that the ionization energy of a hydrogen atom is 13.527eV, calculate the second ionization energy of a helium atom. The general equation for the lines in the emis-sion spectra of one electron system such as H, He^+ , Li^2+ and Be^3+ is given as ѵ= Z^2 R_H{(1/n^2_1) - (1/n^2_2)} whereѵ= the wave numbers of emission lines and Z is the nuclear charge of the atom concerned.

category

chemistry

options

  1. 13.527eV
  2. 162.324eV
  3. 21.654eV
  4. 108.216eV
  5. 40.581eV
  6. 54.108eV
  7. 30.081eV
  8. 72.144eV
  9. 96.192eV
  10. 27.054eV

src

stemez-PhysicalChemistry

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