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(B) The energy of an electron in the $n^{th}$ shell of a hydrogen atom is given by the formula $E_n = -13.6 / n^2 \ eV$.For the $L$ shell,the principa

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$-13.6$

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(B) The energy of an electron in the $n^{th}$ shell of a hydrogen atom is given by the formula $E_n = -13.6 / n^2 \ eV$.For the $L$ shell,the principal quantum number $n = 2$.Therefore,the energy of the electron in the $L$ shell is $E_2 = -13.6 / (2)^2 = -13.6 / 4 = -3.4 \ eV$.When the electron liberates $10.2 \ eV$ of energy,it transitions to a lower energy state.The new energy of the system is $E_{final} = E_{initial} - 	ext{Energy liberated} = -3.4 \ eV - 10.2 \ eV = -13.6 \ eV$.This corresponds to the ground state $(n = 1)$ of the hydrogen atom.

Supposing the electron of the $H$ atom is present in the $L$ shell. If it liberates $10.2 \ eV$,what is the energy of the system? ............. $eV$

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