Question

He atom can be excited to 1s1 2p1 by wavelength 58.44 nm. If lowest excited state for He lies 4857cm-1 below the above. Calculate the energy for the lower excitation state.

Answer 1

Answer:

Ionization energy of helium is to be consider as=2372kJ/mol

IE=$\frac{2372kJ}{mol} *\frac{1000J}{1kJ} *\frac{1mol}{6.023*10^{23}atoms }$

=$3.938*10^{-18} J/atom$

Then energy required to eject the electron from 2p level

x=Ionization energy-energy gap between 2p and 2s

 =$3.938*10^{-18} J/atom-3.4*10^{-18} J/atom$

 =$5.38*10^{-19} J$