Question

The first ionization energy of H and He are 13.6ev and 24.6ev respectively.how much energy would be given out during the formation of ground state of He atom from He +2 nucleus if it combines with two electrons?

Answer 1

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The given data is as follows :
H → H+ + e- ;  IE1 = 13.36 eV 
He → He+ + e-        IE1 = 24.6 eV
We have to determine the values of
He2+ + e- → He+ ;   ΔH = a
He+ + e- → He ;  ΔH = b
The overall energy change can be written as :
He2+ + 2e- → He ;   ΔH = (a + b)

The IE1 of He+(ΔH = a​) = IE1(H) ×Z2
                                               =IE1(H) × 22 
                                       = -13.36 × 4
                                       = -53.44 eV 

The IE1 of He(ΔH = b​) = IE1 = -24.6 eV (given) 
Therefore, total amount of energy given out = a + b
= [– 53.44 + (– 24.6)]eV    
= 78.04 eV

Answer 2

Answer:

The given data is as follows :

H → H+ + e- ;  IE1 = 13.36 eV  

He → He+ + e-        IE1 = 24.6 eV

We have to determine the values of

He2+ + e- → He+ ;   ΔH = a

He+ + e- → He ;  ΔH = b

The overall energy change can be written as :

He2+ + 2e- → He ;   ΔH = (a + b)

The IE1 of He+(ΔH = a​) = IE1(H) ×Z2

                                              =IE1(H) × 22  

                                      = -13.36 × 4

                                      = -53.44 eV  

The IE1 of He(ΔH = b​) = IE1 = -24.6 eV (given)  

Therefore, total amount of energy given out = a + b

= [– 53.44 + (– 24.6)]eV    

= 78.04 eV

Explanation:

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