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A single electron system has ionization energy 11180 kJ `mol^(-1)`. Find the number of protons in the nucleus of the system. |
Answer» Correct Answer - Z=3 `IE=(Z^(2))/(n^(2))xx21.69xx10^(-19)J` `(11180xx10^(3))/(6.023xx10^(23))=(Z^(2))/(1^(2))xx21.69xx10^(-19)` `Z~~3`. |
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