A) 13.6 eV
B) 6.8 eV
C) 54.4 eV
D) 72.2 Ev
Correct Answer: C
Solution :
I.P. for\[H{{e}^{+}}\]ion ==, I.P for\[H\times {{(Z)}^{2}}\] where, Z = atomic number 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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