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The ratio of the enrgy of the electron in groun state of hydrogen to the elecron in first excited state of `Be^(2+)` is :A. ` 1:4`B. `1:8`C. ` 1:16`D. ` 16 :1` |
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Answer» Correct Answer - A `E_(2_((Be^(3+))) = E_(2(H)) xx Z^2` , Also ` E_(2 (H)) = E_(1(H0))/2^2` ` ,. E_(2(Be^(3+0)) = (E_(1(H)))/2^2 xx 4^2 = 4 xx E_(1(H))`. |
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