# Calculate the energy required for the process

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 $^{-1}$

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 $8.72 \times 10^{-18} \mathrm{~J}$.