A doubly charged lithium atom (Li^{++}) is a | vedclass

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

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$113.74$

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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 \frac{Z^2}{n^2} \, eV$.For $Li^{++}$,the atomic number $Z = 3$.The energy required for the transition from $n_1 = 1$ to $n_2 = 3$ is:$\Delta E = E_3 - E_1 = 13.6 	imes Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$$\Delta E = 13.6 	imes 3^2 \left( \frac{1}{1^2} - \frac{1}{3^2} ight) = 13.6 	imes 9 	imes \left( 1 - \frac{1}{9} ight) = 13.6 	imes 9 	imes \frac{8}{9} = 108.8 \, eV$.Using the relation $\Delta E = \frac{hc}{\lambda}$,where $h = 6.626 	imes 10^{-34} \, J \cdot s$,$c = 3 	imes 10^8 \, m/s$,and $1 \, eV = 1.602 	imes 10^{-19} \, J$:$\lambda = \frac{hc}{\Delta E} = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{108.8 	imes 1.602 	imes 10^{-19}} \approx 1.1374 	imes 10^{-8} \, m$.$\lambda = 113.74 \, \mathring{A}$.

$A$ doubly charged lithium atom $(Li^{++})$ is a hydrogen-like species with atomic number $Z = 3$. The wavelength of the radiation required to excite an electron from the first to the third Bohr orbit in $Li^{++}$ is.......$\mathring{A}$ (Ionization energy of hydrogen atom is $13.6\,eV$).

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