Question

Calculate the energy required for the process
$H{e}_{\left(g\right)}^{+}\to H{e}_{\left(g\right)}^{2+}+e^{-}$
The ionization energy for the H atom in the ground state is 2.18 ×${10}^{–18}Jato{m}^{–1}$

Answer 1

Energy associated with hydrogen-like species is given by,
$E_n=-2.18\times {10}^{-18}\left(\frac{Z^2}{n^2}\right)J$
For ground state of hydrogen atom,
$△E=E_{\infty }-E_1=0-[-2.18\times {10}^{-18}\left\{\frac{(1{)}^2}{(1{)}^2}\right\}]J△E=2.18\times {10}^{-18}J$
For the given process,
$H{e}_{\left(g\right)}^{+}\to H{e}_{\left(g\right)}^{2+}+e^{-}$
An electron is removed from n = 1 to n = ∞.
$△E=E_{\infty }-E_1=0-[-2.18\times {10}^{-18}\left\{\frac{(2{)}^2}{(1{)}^2}\right\}]J△E=8.72\times {10}^{-18}J$
$\therefore$ The energy required for the process is $8.72\times {10}^{-18}J$.