NEET-XII-Physics
43: Bohr's Model and Physics of the Atom
- #10A positive ion having just one electron ejects it if a photon of wavelength 228 Å or less is absorbed by it. Identify the ion.Ans : Given:
Wavelength of photon, λ = 228 Å
Energy `` \left(E\right)`` is given by
`` E=\frac{h\,\mathrm{\,c\,}}{\lambda }``
`` ``
Here, c = Speed of light
h = Planck's constant
`` \therefore E=\frac{\left(6.63\times {10}^{-34}\right)\times \left(3\times {10}^{8}\right)}{\left(228\times {10}^{-10}\right)}``
`` =0.0872\times {10}^{-16}\,\mathrm{\,J\,}``
As the transition takes place from n = 1 to n = 2, the excitation energy (E1) will be
`` {E}_{1}=Rh\,\mathrm{\,c\,}{Z}^{2}\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
`` {E}_{1}=\left(13.6\,\mathrm{\,eV\,}\right)\times {Z}^{2}\times \left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)``
`` \Rightarrow {E}_{1}=\left(13.6\,\mathrm{\,eV\,}\right)\times {Z}^{2}\times \frac{3}{4}``
This excitation energy should be equal to the energy of the photon.
`` \therefore 13.6\times \frac{3}{4}\times {\,\mathrm{\,Z\,}}^{2}=0.0872\times {10}^{-16}``
`` {Z}^{2}=\frac{0.0872\times {10}^{-16}\times 4}{13.6\times 3\times 1.6\times {10}^{-19}}=5.34``
`` Z=\sqrt{5.34}=2.3``
The ion may be helium.
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