Question

The ionization energy of hydrogen is $13.6eV$. The first ionization energy of $He$ is $24.6eV$. Calculate the energy given out when $H{e}^{2+}$ nucleus combines with two electrons to form a helium atom in ground state.

• A. $68.2eV$
• B. $51.8eV$
• C. $38.2eV$
• D. $79eV$

Answer 1

Answer: (D)

$79eV$

The first ionization energy of H to $H^{+}$ is 13.6 eV.
The first ionization energy of He to $H{e}^{+}$ is 24.6 eV......(1)
When a $H{e}^{2+}$ ion accepts an electron, the energy released is $Z^2=2^2=4$ times the first ionization energy of hydrogen atom.
Thus, the energy for the ionization of $H{e}^{2+}$ to $H{e}^{+}$ is $4\times 13.6=54.4eV$......(2)
When (1) and (2) are added, we get the energy given out when $H{e}^{2+}$ ion gains 2 electrons and forms He atom in its ground state.
It will be $24.6eV+54.4eV=79eV$.