Energyof anelectron in the groundstate of thehydrogenatom is -2 . 18 xx 10^(-18)J. Calculatetheionizationenthalpyof atomichydrogenin termsof kJ mol^(-1)

Energyof anelectron in the groundstate of thehydrogenatom is -2 . 18 x

1.	Energyof anelectron in the groundstate of thehydrogenatom is -2 . 18 xx 10^(-18)J. Calculatetheionizationenthalpyof atomichydrogenin termsof kJ mol^(-1)
Answer» SOLUTION :Ionizationenergyis theamount ofenergyrequired toremovethe electronfrom thegroundstate to infinity. Now, energy ofthe electronin the groundstate = -`2.18 xx 10^(-15) J` ENERGYOF the electronat affinity=0 The energyrequiredto removedand electronin theground state ofhydrogen atom =0 - (its energyin thegroundstate ) = - `(-2 .18 xx 10^(-18)J )= 2. 18 xx 10^(-18) J` `:.` Ionizationenthalpy per MOLEOF HYDROGENATOMS `= (2.18 xx 10^(18)xx 6.02 xx 10^(23))/(1000) kJ` `= 1312 . 36 kJmol^(-1)= 1312 .36xx 10^(3)J mol^(-1)`