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Q.
A single electron in an ion has ionisation energy equal to $217.6\, eV$. What is the total number of neutrons present in that ion is :
Structure of Atom
Solution:
Ionisation energy of ion $= E _{\infty}- E _{1}=217.6\, eV$
therefore $E _{1}=-217.6\, eV =-13.6 \times \frac{ Z ^{2}}{1^{2}}$
$\therefore Z^{2}=\frac{217.6}{13.6}=16$ and $Z=4$
thus in berylium $( Z =4)$ ion no. of neutrons
$=9-4=5$