If the binding energy of the electron in a hy | vedclass

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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. For a hydrogen atom

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$30.6$

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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. For a hydrogen atom,the binding energy is the energy required to remove the electron from the ground state $(n=1)$,which is $13.6 \ eV$. For $Li^{2+}$,the atomic number $Z = 3$. The first excited state corresponds to $n = 2$. The energy of the electron in the first excited state of $Li^{2+}$ is $E_2 = -13.6 	imes \frac{3^2}{2^2} \ eV = -13.6 	imes \frac{9}{4} \ eV = -30.6 \ eV$. The energy required to remove the electron (binding energy) is the energy needed to take it to infinity ($n = \infty$,where $E = 0$). Therefore,$\Delta E = E_{\infty} - E_2 = 0 - (-30.6 \ eV) = 30.6 \ eV$.

If the binding energy of the electron in a hydrogen atom is $13.6 \ eV$,then the energy required to remove the electron from the first excited state of $Li^{2+}$ is: (in $eV$)

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