Calculate the energy required for the process

\(He^{+}_{(g)}\rightarrow He^{2+}_{(g)}+e^{-}\)

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

Asked by Pragya Singh | 1 year ago |  94

1 Answer

Solution :-

The energy associated with hydrogen-like species is:

\(E_{n}=-2.18\times 10^{-18}(\dfrac{Z^{2}}{n^{2}})J\)

For the ground state of the hydrogen atom,

\(\Delta E=E_{\infty }-E_{1}\\ \\ =0-[-(2.18\times 10^{-18}{\dfrac{(1)^{2}}{(1)^{2}}}]J\\ \\ \Delta E=2.18\times 10^{-18}J\\\)

For the process given by:

\(He^{+}_{(g)}\rightarrow He^{2+}_{(g)}+e^{-}\)

An electron is moved from n = 1 to n = ∞.

 \( \Delta E=E_{\infty }-E_{1}\\ \\ =0-[-(2.18\times 10^{-18}{\dfrac{(2)^{2}}{(1)^{2}}}]J\)

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

The required energy for this process is \( 8.72\times 10^{-18}J\)

Answered by Mayu | 1 year ago

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