NEET-XII-Physics
43: Bohr's Model and Physics of the Atom
- #27A beam of light having wavelengths distributed uniformly between 450 nm to 550 nm passes through a sample of hydrogen gas. Which wavelength will have the least intensity in the transmitted beam?Ans : Given:
Minimum wavelength of the light component present in the beam, `` {\lambda }_{1}`` = 450 nm
Energy associated `` \left({E}_{1}\right)`` with wavelength `` \left({\lambda }_{1}\right)`` is given by
E1 = `` \frac{hc}{{\lambda }_{1}}``
Here,
c = Speed of light
h = Planck's constant
`` \therefore {E}_{1}=\frac{1242}{450}``
= 2.76 eV
Maximum wavelength of the light component present in the beam, `` {\lambda }_{2}`` = 550 nm
Energy associated `` \left({E}_{2}\right)`` with wavelength `` \left({\lambda }_{2}\right)`` is given by
E2 = `` \frac{hc}{{\lambda }_{2}}``
`` \therefore {E}_{2}=\frac{1242}{550}=2.228=2.26\,\mathrm{\,eV\,}``
The given range of wavelengths lies in the visible range.
∴ n1 = 2, n2 = 3, 4, 5 ...
Let E'2 , E'3 , E'4 and E'5 be the energies of the 2nd, 3rd, 4th and 5th states, respectively.
`` E{\text{'}}_{2}-E{\text{'}}_{3}=13.6\left(\frac{1}{4}-\frac{1}{9}\right)``
`` =\frac{12.6\times 5}{30}=1.9\,\mathrm{\,eV\,}``
`` E{\text{'}}_{2}-E{\text{'}}_{4}=13.6\left(\frac{1}{4}-\frac{1}{16}\right)``
`` =2.55\,\mathrm{\,eV\,}``
`` E{\text{'}}_{2}-E{\text{'}}_{5}=13.6\left(\frac{1}{4}-\frac{1}{25}\right)``
`` =\frac{10.5\times 21}{100}=2.856\,\mathrm{\,eV\,}``
Only, E'2 - E'4 comes in the range of the energy provided. So the wavelength of light having 2.55 eV will be absorbed.
`` \,\mathrm{\,\lambda \,}=\frac{1242}{2.55}=487.05\,\mathrm{\,nm\,}``
`` =487\,\mathrm{\,nm\,}``
The wavelength 487 nm will be absorbed by hydrogen gas. So, wavelength ​487 nm will have less intensity in the transmitted beam.
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