Structure of Atom - Result Question 20
####20. If ionization potential for hydrogen atom is $13.6 eV$, then ionization potential for $He^{+}$will be
(a) $54.4 eV$
(b) $6.8 eV$
(c) $13.6 eV$
(d) $24.5 eV$
[1993]
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Solution:
- (a) The ionization energy of any hydrogen like species (having one electron only) is given by the equation
I.E $=\frac{2 \pi^{2} Z^{2} m e^{4}}{h^{2}}$ or $\quad$ I.E $\propto Z^{2}$
Since the atomic number of $H$ is 1 and that of $He$ is 2 , therefore, the I.E. of $He^{+}$is four times $(2^{2})$ the I.E. of Hi.e., $13.6 \times 4=54.4 eV$
For H-like particles,
$E_n=-\frac{21.8 \times 10^{-19}}{h^{2}} Z^{2} J /$ atom
$ =-\frac{1312}{n^{2}} Z^{2} kJ / mol $
$I . E=E _{a_o}-E_1=0-(-I \cdot E_H Z^{2})$
$=Z^{2} \times I_E H$
Now for $He^{+}, Z=2 \quad \therefore \quad I . E=4 \times I_H$
$ \text{ for } Li^{2+}, Z=3 \quad \therefore \quad I . E=9 \times I_H $
