Question

Energy required for the process is :

${He}^{+}g⟶H{e}^{+2}\left(g\right)+e^{-}$

(The ionization energy for the H-atom in the ground state is $2.18\times {10}^{18}J{atom}^{-1}$)

• A. $9.12\times {10}^{-18}J$
• B. $8.72\times {10}^{-18}J$
• C. $4.72\times {10}^{-20}J$
• D. $7.82\times {10}^{-16}J$

Answer 1

The correct option is B $8.72\times {10}^{-18}J$

$E_n=-2.18\times {10}^{-18}\left[\frac{Z^2}{n^2}\right]$

$\Delta E=E_{\infty }-E_1=0-[-2.18\times {10}^{-18}\left[\frac{(1{)}^2}{(1{)}^2}\right]]$ $J$

$\Delta E=2.18\times {10}^{-18}$ $J$

For $H{e}^{+}⟶H{e}^{2+}+e^{-}$

An electron is removed by $n=1$ to $n=\infty$

$\Delta E=E_{\infty }-E_1=0-[-2.18\times {10}^{-18}\left[\frac{(2{)}^2}{(1{)}^2}\right]]$ $J$

$\Delta E=8.72\times {10}^{-18}$ $J$

$\therefore$ The energy required for the process is $8.72\times {10}^{-18}$ $J$