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Ionisation energy of `He^+` is `19.6 xx 10^-18 J "atom"^(-1)`. The energy of the first stationary state `(n = 1)` of `Li^( 2 +)` is.A. `8.82xx010^(-17) J "atom"^(-1)`B. `4.41xx10^(-16) J "atom"^(-1)`C. `-4.41xx10^(-17) J "atom"^(-1)`D. `-2.2xx10^(-15) J "atom"^(-1)` |
Answer» Correct Answer - C `IE=-E_(1)` `E_(1) "for" He^(+) "si" =-19.6xx10^(-18) "J mol"^(-1)` `((E_(1))_(He^(+)))/((E_(1))_(Li^(2+)))=((Z_(He^(+)))^(3))/((Z_(Li^(2+)))^(2))` `(-19.6xx10^(-18))/((E_(1))_(Li^(2+)))=(4)/(9)` `E_(1_(Li^(2+)))=((-19.6xx10^(-18))xx9)/(4)` `=-4.41 "J atom"^(-1)` |
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