Ionization energy of He+ ion at minimum energy position is

Question

Ionization energy of He+ ion at minimum energy position is (A) 13.6 eV (B) 27.2 eV (C) 54.4 eV (D) 68.0 eV

Tardigrade Admin · Accepted Answer

The energy of helium ions. En=-(13 .6 Z2/n2) eV In minimum position, n=1 . For He+, Z=2 . E=(-13.6 ×(2)2/1) eV E=- text54.4 eV Ionization Energy = -E=54.4eV

Q. Ionization energy of $H e^{+}$ ion at minimum energy position is

$13.6 e V$

$27.2 e V$

$54.4 e V$

$68.0 e V$

The energy of helium ions.
$E_{n} = - \frac{13.6 Z ^{2}}{n ^{2}} e V$
In minimum position, $n = 1$ .
For $H e^{+}, Z = 2$ .
$E = \frac{- 13.6 \times ( 2 ) ^{2}}{1} e V$
$E = - 54.4 e V$
Ionization Energy = $- E = 54.4 e V$

Q. Ionization energy of He+ ion at minimum energy position is

13.6eV

27.2eV

54.4eV

68.0eV