A) 13.6 eV
B) 27.2 eV
C) 54.4 eV
D) 68.0 eV
Correct Answer: A
Solution :
Energy of helium ions. \[{{E}_{n}}=-\frac{13.6{{Z}^{2}}}{{{n}^{2}}}eV\] In minimum position,\[n=1\] For\[H{{e}^{2+}},\] \[Z=2\] \[E=\frac{-13.6\times {{(2)}^{2}}}{1}eV\] \[E=54.4eV\]You need to login to perform this action.
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