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(B) The energy of the incident photon is given by $E_{	ext{photon}} = \frac{hc}{\lambda} = \frac{1242 \ eV \ nm}{90 \ nm} = 13.8 \ eV$.The energy req

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$2$

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(B) The energy of the incident photon is given by $E_{	ext{photon}} = \frac{hc}{\lambda} = \frac{1242 \ eV \ nm}{90 \ nm} = 13.8 \ eV$.The energy required to ionize a hydrogen atom from the $n^{	ext{th}}$ orbital is $E_n = \frac{13.6 \ eV}{n^2}$.According to the law of conservation of energy,the energy of the photon is equal to the sum of the ionization energy and the kinetic energy of the ejected electron:$E_{	ext{photon}} = E_n + K.E.$Substituting the given values:$13.8 \ eV = \frac{13.6 \ eV}{n^2} + 10.4 \ eV$.Rearranging the equation:$\frac{13.6}{n^2} = 13.8 - 10.4 = 3.4$.Solving for $n^2$:$n^2 = \frac{13.6}{3.4} = 4$.Therefore,$n = 2$.

Consider a hydrogen atom with its electron in the $n^{\text{th}}$ orbital. An electromagnetic radiation of wavelength $90 \ nm$ is used to ionize the atom. If the kinetic energy of the ejected electron is $10.4 \ eV$,then the value of $n$ is $(hc = 1242 \ eV \ nm)$.

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