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2.34    Calculate the energy required for the process

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

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

Answers (1)

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For the hydrogen-like particles,

E_{n} =-\frac{2n^2mZ^2e^4}{n^2h^2}

For H-atom, Ionization energy:

I.E. =E-E_{1} = 0-(-\frac{2\pi^2me^4}{1^2h^2})= \frac{2\pi^2me^4}{h^2}

= 2.18\times10^{-18}J/atom       (Given)

For the given process, the energy required will be:

E_{n} - E_{1}

= 0-(-\frac{2\pi^2me^4}{1^2h^2})

= 4\times\frac{2\pi^2me^4}{h^2}

= 4\times2.18\times10^{-18}J

= 8.72\times10^{-18}J

Posted by

Divya Prakash Singh

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