If 13.6 eV energy is required to ionize the | vedclass

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(B) The energy of an electron in the $n$th orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \ eV$.To remove an electron from

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$3.4$

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(B) The energy of an electron in the $n$th orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \ eV$.To remove an electron from the $n=2$ orbit to infinity (ionization),we need to provide energy equal to the magnitude of the binding energy at that level.The energy of the electron in the $n=2$ state is $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \ eV$.The energy required to remove this electron is the difference between the energy at infinity $(0 \ eV)$ and the energy at $n=2$ $(E_{	ext{req}} = 0 - (-3.4) = 3.4 \ eV)$.

If $13.6 \ eV$ energy is required to ionize the hydrogen atom,then the energy required to remove an electron from $n=2$ is (in $eV$)

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