Question

The ionization potential of an electron in the second orbit of $H{e}^{+}$ atom is

• A. 13.6 eV
• B. 27.2 eV
• C. 54.4 eV
• D. 100 eV

Answer 1

The correct option is C 54.4 eV

For the ionization of second He electron. $H{e}^{+}$ will act as hydrogen like atom.
Hence ionization potential
$=Z^2\times 13.6volt=(2{)}^2\times 13.6=54.4V$