A He+ ion is in its first excited state. Its ionization energy is :

Question

A He+ ion is in its first excited state. Its ionization energy is : (A) 6.04 eV (B) 13.60 eV (C) 54.40 eV (D) 48.36 eV

Tardigrade Admin · Accepted Answer

E = -13.6 (z2/n2) eV As He+ is 1st excited state ∴ z = 2, n = 2 E = -13.6 eV As total energy of He+ in 1st excited state is -13.6 eV, ionisation energy should be +13.6 eV.

Q. A $H e^{+}$ ion is in its first excited state. Its ionization energy is :

$6.04 e V$

$13.60 e V$

$54.40 e V$

$48.36 e V$

$E = - 13.6 \frac{z ^{2}}{n ^{2}} e V$
As He+ is 1st excited state
$∴ z = 2, n = 2$
$E = - 13.6 e V$
As total energy of $H e^{+}$ in 1st excited state is $- 13.6 e V$ , ionisation energy should be $+ 13.6 e V$ .

Q. A He+ ion is in its first excited state. Its ionization energy is :

6.04eV

13.60eV

54.40eV

48.36eV