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A neutral atom (Ar) is converted to (Ar^{+3}) by the following process

$\mathrm{Ar}\underset{-\mathrm{r}}{\overset{{\mathrm{E}}_{1}}{\to}}{\mathrm{Ar}}^{+}\underset{-\mathrm{r}}{\overset{{\mathrm{E}}_{2}}{\to}}{\mathrm{Ar}}^{2+}\underset{-\mathrm{r}}{\overset{{\mathrm{E}}_{3}}{\to}}{\mathrm{Ar}}^{3+}$

The correct order of E_{1}, E_{2} and E_{3} _{ }energies is

(1) E_{1} < E_{2} < E_{3} _{ }(2) E_{1} > E_{2} > E_{3 }(3) E_{1} = E_{2} = E_{3 }(4) E_{1} > E_{2} < E_{3}

The Ar ionization energy is the energy required to remoove from atom one mole of electrons with subsequent production of positivity charged ion of Argon.

$Ar\to A{r}^{+}+{e}^{-}\phantom{\rule{0ex}{0ex}}$

This process can be repeated many times and so energy cost also increased automatically. The general equation for this process will be-

$A{r}^{n+}\to A{r}^{(n+1)}+e$

so for the above question:

**option 1 is correct**

${E}_{1}<{E}_{2}<{E}_{3}$

${E}_{1}<{E}_{2}<{E}_{3}$

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