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Calculate the energy required for the process
\(\mathrm{He}^{+}(\mathrm{g}) \longrightarrow \mathrm{He}^{2+}(\mathrm{g})+\mathrm{e}^{-}\)
The ionization energy for the \(\mathbf{H}\) atom in the ground state is \(2.18 \times 10^{-18} \mathrm{~J}^2\) atom \({ }^{-1}\).
\(\mathrm{He}^{+}(\mathrm{g}) \longrightarrow \mathrm{He}^{2+}(\mathrm{g})+\mathrm{e}^{-}\)
The ionization energy for the \(\mathbf{H}\) atom in the ground state is \(2.18 \times 10^{-18} \mathrm{~J}^2\) atom \({ }^{-1}\).
Solution:
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Verified Answer
For H-like particles, \(E_n=\frac{2 \pi^2 m Z^2 e^4}{n^2 h^2}\)
For H-atom, I.E. \(=E-E_1=0-\left(-\frac{2 \pi^2 m e^4}{1^2 \times h^2}\right)\)
\(=\frac{2 \pi^2 m e^4}{h^2}=2.18 \times 10^{-18} J_{\text {atoms }^{-1} \text { (Given) }}\)
For the given process,
\(\begin{aligned}
&\text { Energy required }=E_{\mathrm{n}}-E_1=0-\left(-\frac{2 \pi^2 m \times 2^2 \times e^4}{l^2 \times h^2}\right) \\
&=4 \times \frac{2 \pi^2 m e^4}{h^2} \\
&=4 \times 2.18 \times 10^{-18}=8.72 \times 10^{-18} \mathrm{~J}
\end{aligned}\)
For H-atom, I.E. \(=E-E_1=0-\left(-\frac{2 \pi^2 m e^4}{1^2 \times h^2}\right)\)
\(=\frac{2 \pi^2 m e^4}{h^2}=2.18 \times 10^{-18} J_{\text {atoms }^{-1} \text { (Given) }}\)
For the given process,
\(\begin{aligned}
&\text { Energy required }=E_{\mathrm{n}}-E_1=0-\left(-\frac{2 \pi^2 m \times 2^2 \times e^4}{l^2 \times h^2}\right) \\
&=4 \times \frac{2 \pi^2 m e^4}{h^2} \\
&=4 \times 2.18 \times 10^{-18}=8.72 \times 10^{-18} \mathrm{~J}
\end{aligned}\)
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