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Q.

Hydrogen atom with its electron in the nth orbit is ionized by electromagnetic radiation of wavelength 45 nm. If the kinetic energy of the ejected electron is 26.75 eV then the value of n is (hc = 1242 eV nm)

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answer is 4.

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Detailed Solution

$E = \frac{1242}{45} e V = 27.6 e V 27.6 - 26.75 = 0.85 e V 0.85 = \frac{13.6}{n^{2}} \Rightarrow n^{2} = 16 \Rightarrow n = 4$

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Hydrogen atom with its electron in the nth orbit is ionized by electromagnetic radiation of wavelength 45 nm. If the kinetic energy of the ejected electron is 26.75 eV then the value of n is (hc = 1242 eV nm)