Loading…
NEET-XII-Physics

43: Bohr's Model and Physics of the Atom

with Solutions - page 4
Qstn# iv-12-a Prvs-QstnNext-Qstn
  • #12-a
    Identify the quantum numbers n of the upper and the lower energy states involved in the transition. (b) Find the wavelength of the emitted radiation. (b) Find the wavelength of the emitted radiation.
    Ans : The binding energy of hydrogen is given by
    `` E=\frac{13.6}{{n}^{2}}\,\mathrm{\,eV\,}``
    For binding energy of 0.85 eV,
    `` {{n}_{2}}^{2}=\frac{13.6}{0.85}=16``
    `` {n}_{2}=4``
    For binding energy of 10.2 eV,
    `` {{n}_{1}}^{2}=\frac{13.6}{10.2}``
    `` {n}_{1}=1.15``
    `` \Rightarrow {n}_{1}=2``
    `` ``
    The quantum number of the upper and the lower energy state are 4 and 2, respectively. (b) Wavelength of the emitted radiation `` \left(\lambda \right)`` is given by
    `` \frac{1}{\lambda }=R\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
    Here,
    R = Rydberg constant
    n1 and n2 are quantum numbers.
    `` \therefore \frac{1}{\lambda }=1.097\times {10}^{7}\left(\frac{1}{4}-\frac{1}{16}\right)``
    `` \Rightarrow \,\mathrm{\,\lambda \,}=\frac{16}{1.097\times 3\times {10}^{7}}``
    `` =4.8617\times {10}^{-7}``
    `` =487\,\mathrm{\,nm\,}``
    Page No 384: (b) Wavelength of the emitted radiation `` \left(\lambda \right)`` is given by
    `` \frac{1}{\lambda }=R\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
    Here,
    R = Rydberg constant
    n1 and n2 are quantum numbers.
    `` \therefore \frac{1}{\lambda }=1.097\times {10}^{7}\left(\frac{1}{4}-\frac{1}{16}\right)``
    `` \Rightarrow \,\mathrm{\,\lambda \,}=\frac{16}{1.097\times 3\times {10}^{7}}``
    `` =4.8617\times {10}^{-7}``
    `` =487\,\mathrm{\,nm\,}``
    Page No 384:
  • #12-b
    Find the wavelength of the emitted radiation.
    Ans : Wavelength of the emitted radiation `` \left(\lambda \right)`` is given by
    `` \frac{1}{\lambda }=R\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
    Here,
    R = Rydberg constant
    n1 and n2 are quantum numbers.
    `` \therefore \frac{1}{\lambda }=1.097\times {10}^{7}\left(\frac{1}{4}-\frac{1}{16}\right)``
    `` \Rightarrow \,\mathrm{\,\lambda \,}=\frac{16}{1.097\times 3\times {10}^{7}}``
    `` =4.8617\times {10}^{-7}``
    `` =487\,\mathrm{\,nm\,}``
    Page No 384: