Structure Of Atom Ques 13
13. If ionization potential for hydrogen atom is $13.6$ $ eV$, then ionization potential for $He^{+}$will be
[1993]
(a) $54.4$ $ eV$
(b) $6.8$ $ eV$
(c) $13.6$ $ eV$
(d) $24.5$ $ eV$
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Answer:
Correct Answer: 13.(a)
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}}\quad $ 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 $H$ i.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 $
BETA
