Consider a hydrogen atom with its electron in the nth orbit. An electromagnetic radiation of wavelength 90nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4eV, then the value of n is (take hc=1242eV-nm)

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

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Detailed Solution

$E_{inc} (eV) = \frac{1242 eV-nm}{λ (nm)} = \frac{1242}{90} = 13.8 eV$

Energy of the electron $= - \frac{13.6 eV}{n^{2}} + 13.8 = 0 + 10.4 eV$

$= - \frac{13.6 e V}{n^{2}} = - 3.4 e V$

$= n^{2} = \frac{13.6 e V}{3.4 e V} = 4 \Rightarrow n = 2$

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