JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 6)
A He+ ion is in its first excited state. Its
ionization energy is :-
13.60 eV
6.04 eV
48.36 eV
54.40 eV
Explanation
$$T.E. = - \left( {13.6} \right)\left( {{{{Z^2}} \over {{n^2}}}} \right)eV$$
z = n = 2
$$ \Rightarrow $$ Ionisation energy = –T.E. = 13.6 eV
z = n = 2
$$ \Rightarrow $$ Ionisation energy = –T.E. = 13.6 eV
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