The ionization enthalpy of hydrogen atom...

The ionization enthalpy of hydrogen atoms is `1.312xx10^(6) J" mol"^(-1)` The energy required to excite the electron the atom from n =1 to n=2 is :

`9.84xx10^(5) J" mol"^(-1)`

`8.51xx10^(6)J" mol"^(-1)`

`6.56xx10^(5)J" mol"^(-1)`

`7.56xx10^(5) J" mol"^(-1)`

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The correct Answer is:

`E_(1)=-1.312xx10^(6)" J mol"^(-1), E_(2)=(E_(1))/(2^(2))=(1.312xx10^(6))/(4)" J mol"^(-1)`
`Delta E =(E_(2)-E_(1)) =1.312 xx10^(6) (1-(1)/(4))=(3)/(4)xx1.312xx10^(6)=9.84xx10^(5)" J mol"^(-1)`

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