Question

The ionization potential for hydrogen atom is $13.6$ eV. The ionization energy for $H{e}^{+}$ ion should be :

• A. $27.2$ eV
• B. $54.4$ eV
• C. $6.8$ eV
• D. $3.4$ eV

Answer 1

Answer: (B)

$54.4$ eV

The ionization potential for $H{e}^{+}$ ion is $54.4eV$.

Bohr's model can be used to calculate ionization energy of only "Hydrogen like" ions.
$H{e}^{+}$ is an hydrogen like ion.

$E=-\frac{13.6z^2}{n^2}$ can be used to approximate the Ionization potential with $z=2$ and $n=1$ ($H{e}^{+}$ is $1s^1$ configuration).

$H{e}^{+}=54.4eV$

Note that here we are calculating the $2^{nd}$ ionization energy i.e., $H{e}^{+}$ to $H{e}^{++}$.