JEE Advance - Physics (2015 - Paper 1 Offline - No. 10)
Consider a hydrogen atom with its electron in the nth 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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Објашњење
Energy of the incident photon $$ = hf = {{hc} \over \lambda } = {{1242} \over {90}} = 13.8$$ eV. Since after ionisation, electron is ejected with some kinetic energy. By energy conservation, we get
Energy (photon) = Kinetic energy (electron) + $$\Delta$$E
Transition energy from nth orbit to n $$\to$$ $$\infty$$. Therefore,
13.8 = 10.4 + $$\Delta$$E
$$\Rightarrow$$ $$\Delta$$E = 3.4 eV
From Bohr's theory,
$${E_n} = {{ - 13.6} \over {{n^2}}} = - 3.4 \Rightarrow n = 2$$