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The ratio of ionization energy of `H` and `Be^(+3)` is.A. `1 : 1`B. `1 : 3`C. `1 : 9`D. `1 : 16` |
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Answer» Correct Answer - D (d) `IE = E_(prop) - E_n` `IE prop Z_2` So, `(IE_H)/(IE_(Be^(_3))) = (1)/(16)` `:. "ratio is" 1 : 16`. |
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