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What will be the difference of Z("eff") ...

What will be the difference of `Z_("eff")` in `._(11)Na` and `._(19)K` for last electron -

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The correct Answer is:
0

`Z_("eff") = Z - sigma`
for `Na, K, Rb, Cs Z_("eff") = 2.2`
Thus `Z_("eff")` of `NA - Z_("eff")` of `K = 0`
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