A) \[13.6eV\]
B) \[6.8eV\]
C) \[54.4eV\]
D) \[72.2eV\]
Correct Answer: C
Solution :
I.P for \[H{{e}^{+}}\] ion = I.P for \[H\times {{(Z)}^{2}}\]where, Z = atomic number \[\therefore \] \[I.P.=13.6\times {{(2)}^{2}}\] \[13.6\times 4=54.4eV\]You need to login to perform this action.
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