The one electron species having ionization energy of 54.4 eV is

Question

The one electron species having ionization energy of 54.4 eV is (A)  H  (B)  He+  (C)  B4+  (D)  Li2+

Tardigrade Admin · Accepted Answer

En=(-13.6× (Z)2/n2)eV If n=1 En=-13.6 Z2 ∴ Z2=(- 54.4/-13.6) Z2=4 Z=2 Thus, the species with atomic number 2 (i.e., He+ ) has the ionisation energy 54.4 eV.

Q. The one electron species having ionization energy of 54.4 eV is

$H$

$H e^{+}$

$B^{4 +}$

$L i^{2 +}$

$E_{n} = \frac{- 13.6 \times ( Z ) ^{2}}{n ^{2}} e V$ If $n = 1$ $E_{n} = - 13.6 Z^{2}$ $∴$ $Z^{2} = \frac{- 54.4}{- 13.6}$ $Z^{2} = 4$ $Z = 2$ Thus, the species with atomic number 2 (i.e., $H e^{+}$ ) has the ionisation energy 54.4 eV.

Q. The one electron species having ionization energy of 54.4 eV is

H

He+

B4+

Li2+

En​=n2−13.6×(Z)2​eV If n=1 En​=−13.6Z2 ∴ Z2=−13.6−54.4​ Z2=4 Z=2 Thus, the species with atomic number 2 (i.e., He+ ) has the ionisation energy 54.4 eV.

$H$

$H e^{+}$

$B^{4 +}$

$L i^{2 +}$

$E_{n} = \frac{- 13.6 \times ( Z ) ^{2}}{n ^{2}} e V$ If $n = 1$ $E_{n} = - 13.6 Z^{2}$ $∴$ $Z^{2} = \frac{- 54.4}{- 13.6}$ $Z^{2} = 4$ $Z = 2$ Thus, the species with atomic number 2 (i.e., $H e^{+}$ ) has the ionisation energy 54.4 eV.