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(C) The energy of an electron in the $n^{th}$ shell of a hydrogen atom is given by $E_n = -\frac{E_1}{n^2}$,where $E_1 = 2.18 	imes 10^{-18} \ J$ is

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$2.42 	imes 10^{-19} \ J$

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(C) The energy of an electron in the $n^{th}$ shell of a hydrogen atom is given by $E_n = -\frac{E_1}{n^2}$,where $E_1 = 2.18 	imes 10^{-18} \ J$ is the ionization energy of the ground state $(n=1)$.For $n = 3$,the energy of the electron is $E_3 = -\frac{2.18 	imes 10^{-18}}{3^2} \ J$.$E_3 = -\frac{2.18 	imes 10^{-18}}{9} \ J$.To ionize the atom,we must provide energy equal to the magnitude of the electron's energy in that state to bring it to $E = 0$.Required Energy $= |E_3| = \frac{2.18 	imes 10^{-18}}{9} \ J$.Required Energy $= 0.2422 	imes 10^{-18} \ J = 2.42 	imes 10^{-19} \ J$.

If the electron in a hydrogen atom were in the energy level with $n = 3$,how much energy in joule would be required to ionise the atom? (Ionisation energy of $H$-atom is $2.18 \times 10^{-18} \ J$):

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