A) \[51.8eV\]
B) \[79\text{ }eV\]
C) \[38.2eV\]
D) \[49.2\text{ }eV\]
Correct Answer: B
Solution :
: First ionization energy of helium atom is \[{{I}_{1}}=24.6eV\] Second ionization energy of helium atom is \[{{I}_{2}}=\frac{13.6{{(2)}^{2}}}{{{(1)}^{2}}}eV=54.4eV\] \[\therefore \] Energy required to remove both the electrons from helium atom is \[={{I}_{1}}+{{I}_{2}}=24.6eV+54.4eV=79eV\]You need to login to perform this action.
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