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The ionization energy of He^(+) is 8.72 ...

The ionization energy of `He^(+)` is `8.72 xx 10^(-18)J "atom"^(-1)`. Calculate the energy of first stationary state of `Li^(2+)`.

Text Solution

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`E_(n) = - (2pi^(2) mZ^(2) e^(4))/(n^(2) h^(2)) = - K (Z^(2))/(n^(2))` (K = constant)
I.E. of `He^(+) = E_(oo) - E_(1) = 0 - (-K (2^(2))/(1^(2)))= 4K`
Hence, `4K = 8.72 xx 10^(-18)J "atom"^(-1)` (Given) or `K = 2.18 xx 10^(-18) J " atom"^(-1)`
For `Li^(2+), Z =3` and for 1st stationary state, n = 1
`:. E_(1) = -K (Z^(2))/(n^(2)) = - 2.18 xx 10^(-18) xx (3^(2))/(1^(2)) = - 19.62 xx 10^(-18) J "atom"^(-1)`
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