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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \ eV$.For the ground state,$n

Vedclass · Accepted Answer

$1.51$

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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \ eV$.For the ground state,$n = 1$.The first excited state corresponds to $n = 2$.The second excited state corresponds to $n = 3$.Substituting $n = 3$ into the formula,we get $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \ eV$.To release the electron (ionize the atom),we need to provide enough energy to bring its total energy to $0 \ eV$.Therefore,the energy required is $E_{	ext{ionization}} = 0 - E_3 = 0 - (-1.51 \ eV) = 1.51 \ eV$.

How much energy is needed to release the electron from the second excited state in the hydrogen atom (in $eV$)?

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