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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 :

A

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

B

`8.51xx10^(5)"J mol"^(-1)`

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
`E_(1)= -1.312 xx 10^(6)"J mol"^(-1), E_(2)= (E_(1))/(2^(2))= -(1.312xx10^(6))/(4)"J mol"^(-1)`
`DeltaE= (E_(2)-E_(1))1.312 xx 10^(6)(1-(1)/(4)) =(3)/(4) xx 1.312 xx 10^(6) =9.84 xx 10^(5)"J mol"^(-1)`
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