In: Physics
Part A
Suppose an electron is in the n=6 energy level of a H atom.
How much higher in energy is this electron with respect to the ground state of H
Part B
How much additional energy is required to just remove this excited electron from the H atom?
1) according bore's formula for energy
E =RH[(1/ni2) - (1/nf2)]
here RH is constant =2.179 x10-18 J
ni =intial level =1 and nf =final level =6
E =(2.179 x10-18 J )[(1/1) -(1/36)] = 2.1185 x10-18 J =2.1185 x10-18 x(6.241509·1018 eV) =13.22 eV
energy difference is 13.22 eV
2)additional energy required to remove the electron is ionization energy
E =-Z2RH (1/n2)
here Z= atomic number of hydrogen =1 and n=6
E =-2.179 x10-18 J (1/36) =6.0528 x10-20 J =6.0528 x10-20 x6.241509·1018 eV) =0.38 eV