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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+)` isA. `-2.2 xx 10^(-15) J "atom"^(-1)`B. `8.82 xx 10^(-17) J "atom"^(-1)`C. `4.41 xx 10^(-16)J "atom"^(-1)`D. `-4.41 xx 10^(-17) J "atom"^(-1)` |
Answer» Correct Answer - D `E_(n) = - (2pi^(2) m Z^(2) e^(4))/(n^(2) h^(2)) = - K (Z^(2))/(n^(2))` I.E. of `H^(+) = E_(oo) - E_(1) = 0 - (-K (2^(2))/(1^(2)))` `= 4K = 19.6 xx 10^(-18) J "atom"^(-1)` (Given) or `K = 4.9 xx 10^(-18) J "atom"^(-1)` For `Li^(2+) , Z = 3` and for first stationary state, n = 1 `:. E_(1) = - K (Z^(2))/(n^(2)) = - 4.9 xx 10^(-18) xx (3^(2))/(1^(2))` `= -4.41 xx 10^(-17) J "atom"^(-1)` |
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