The ratio of ionization energy of Bohr's | vedclass

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Question

Energy of an electron in ground state of an atom (Bohr's hydrogen like atom) is given as

$\mathrm{E}=-13.6 \mathrm{Z}^{2} \mathrm{\,eV}$ ( $\ma

Vedclass · Accepted Answer

$1 : 9$

Vedclass · Answer

Energy of an electron in ground state of an atom (Bohr's hydrogen like atom) is given as

$\mathrm{E}=-13.6 \mathrm{Z}^{2} \mathrm{\,eV}$ ( $\mathrm{Z}=$ atomic number of the atom)

$\Rightarrow \mathrm{E}_{	ext {ionisation }}=13.6 \mathrm{\,Z}^{2}$

$\Rightarrow \frac{\left(\mathrm{E}_{\mathrm{ion}}ight)_{\mathrm{H}}}{\left(\mathrm{E}_{\mathrm{ion}}ight)_{\mathrm{Li}}}=\left(\frac{\mathrm{Z}_{\mathrm{H}}}{\mathrm{Z}_{\mathrm{Li}}}ight)^{2}=\left(\frac{1}{3}ight)^{2}=\frac{1}{9}$

The ratio of ionization energy of Bohr's hydrogen atom and Bohr's hydrogen like lithium atom is

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