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As per Bohr model, the minimum energy (in eV) required to remove and electron from the ground state of double ionized Li atom $(Z=3)$ is
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Verified Answer
The correct answer is:
$122.4$
$$
\begin{array}{r}
E=-2^{2} \times 13.6 \mathrm{eV} \\
\qquad \begin{aligned}
E &=-9 \times 1.6 \mathrm{eV} \\
E &=-122.4 \mathrm{eV}
\end{aligned}
\end{array}
$$
So, ionization energy is $+122.4 \mathrm{eV}$
\begin{array}{r}
E=-2^{2} \times 13.6 \mathrm{eV} \\
\qquad \begin{aligned}
E &=-9 \times 1.6 \mathrm{eV} \\
E &=-122.4 \mathrm{eV}
\end{aligned}
\end{array}
$$
So, ionization energy is $+122.4 \mathrm{eV}$
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