Question.
Calculate the energy required for the process
$\mathrm{He}_{(\mathrm{g})}^{+} \rightarrow \mathrm{He}^{2+}{ }_{(\mathrm{g})}+\mathrm{e}^{-}$
The ionization energy for the $\mathrm{H}$ atom in the ground state is $2.18 \times 10^{-18} \mathrm{~J}$ atom $^{-1}$
Calculate the energy required for the process
$\mathrm{He}_{(\mathrm{g})}^{+} \rightarrow \mathrm{He}^{2+}{ }_{(\mathrm{g})}+\mathrm{e}^{-}$
The ionization energy for the $\mathrm{H}$ atom in the ground state is $2.18 \times 10^{-18} \mathrm{~J}$ atom $^{-1}$
Solution:
Energy associated with hydrogen-like species is given by,
$E_{n}=-2.18 \times 10^{-18}\left(\frac{Z^{2}}{n^{2}}\right) \mathrm{J}$
For ground state of hydrogen atom,
$\Delta E=E_{\infty}-E_{1}$
$=0-\left[-2.18 \times 10^{-18}\left\{\frac{(1)^{2}}{(1)^{2}}\right\}\right] J$
$\Delta E=2.18 \times 10^{-18} \mathrm{~J}$
For the given process,
$\mathrm{He}_{(g)}^{+} \rightarrow \mathrm{He}^{2+}(g)+\mathrm{e}^{-}$
An electron is removed from $n=1$ to $n=\infty$.
$\Delta E=E_{\infty}-E_{1}$
$=0-\left[-2.18 \times 10^{-18}\left\{\frac{(2)^{2}}{(1)^{2}}\right\}\right]$
$\Delta E=8.72 \times 10^{-18} \mathrm{~J}$
$\therefore$ The energy required for the process is $8.72 \times 10^{-18} \mathrm{~J}$.
Energy associated with hydrogen-like species is given by,
$E_{n}=-2.18 \times 10^{-18}\left(\frac{Z^{2}}{n^{2}}\right) \mathrm{J}$
For ground state of hydrogen atom,
$\Delta E=E_{\infty}-E_{1}$
$=0-\left[-2.18 \times 10^{-18}\left\{\frac{(1)^{2}}{(1)^{2}}\right\}\right] J$
$\Delta E=2.18 \times 10^{-18} \mathrm{~J}$
For the given process,
$\mathrm{He}_{(g)}^{+} \rightarrow \mathrm{He}^{2+}(g)+\mathrm{e}^{-}$
An electron is removed from $n=1$ to $n=\infty$.
$\Delta E=E_{\infty}-E_{1}$
$=0-\left[-2.18 \times 10^{-18}\left\{\frac{(2)^{2}}{(1)^{2}}\right\}\right]$
$\Delta E=8.72 \times 10^{-18} \mathrm{~J}$
$\therefore$ The energy required for the process is $8.72 \times 10^{-18} \mathrm{~J}$.
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