The energy required to knock out the electron | vedclass

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Question

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.To knock out the electron (i.e.,to i

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$+\frac{13.6}{9} \ eV$

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(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.To knock out the electron (i.e.,to ionize it),we need to provide energy equal to the magnitude of the binding energy of that orbit.The energy required to remove the electron from the $n^{th}$ orbit is $E = +\frac{13.6}{n^2} \ eV$.For the third orbit,$n = 3$.Therefore,the energy required is $E = +\frac{13.6}{3^2} \ eV = +\frac{13.6}{9} \ eV$.

The energy required to knock out the electron in the third orbit of a hydrogen atom is equal to

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