Bohr's atom model

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Question

Find the ratio of ionization energy of hydrogen-like lithium atom and Bohr's hydrogen atom.

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Solution

Energy of an electron in ground state of Bohr's hydrogen-like atom is given by,
$E = - \frac{13.6 Z^{2}}{n^{2}} eV$
where Z = atomic number of the atom.
The ionization energy of this atom is equal in magnitude to energy of ground state $= E_{\infty} = 13.6 Z^{2}$
$\Rightarrow \frac{{(E_{\infty})}_{Li}}{{(E_{\infty})}_{H}} = \frac{{(Z_{Li})}^{2}}{{(Z_{H})}^{2}} = {(\frac{3}{1})}^{2} = 9$

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Bohr's atom model

Remember concepts with our Masterclasses.

Find the ratio of ionization energy of hydrogen-like lithium atom and Bohr's hydrogen atom.

Energy of an electron in ground state of Bohr's hydrogen-like atom is given by,E=−13.6Z2n2eVwhere Z = atomic number of the atom.The ionization energy of this atom is equal in magnitude to energy of ground state=E∞=13.6Z2⇒E∞LiE∞H=ZLi2ZH2=312=9

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