NEET-XII-Physics

43: Bohr's Model and Physics of the Atom

with Solutions -
  • #1
    How many wavelengths are emitted by atomic hydrogen in visible range (380 nm - 780 nm)? In the range 50 nm to 100 nm?
    Ans : Balmer series contains wavelengths ranging from 364 nm (for n2 = 3) to 655 nm (n2 = `` \infty ``).
    So, the given range of wavelength (380-780 nm) lies in the Balmer series.
    The wavelength in the Balmer series can be found by
    `` \frac{1}{\lambda }=R\left(\frac{1}{{2}^{2}}-\frac{1}{{n}^{2}}\right)``
    Here, R = Rydberg's constant = 1.097×107 m`` -``1
    The wavelength for the transition from n = 3 to n = 2 is given by
    `` \frac{1}{{\lambda }_{1}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{3}^{2}}\right)``
    `` {\lambda }_{1}=656.3\,\mathrm{\,nm\,}``
    The wavelength for the transition from n = 4 to n = 2 is given by
    `` \frac{1}{{\lambda }_{2}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{4}^{2}}\right)``
    `` {\lambda }_{2}=486.1\,\mathrm{\,nm\,}``
    The wavelength for the transition from n = 5 to n = 2 is given by
    `` \frac{1}{{\lambda }_{3}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{5}^{2}}\right)``
    `` {\lambda }_{3}=434.0\,\mathrm{\,nm\,}``
    The wavelength for the transition from n = 6 to n = 2 is given by
    `` \frac{1}{{\lambda }_{4}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{6}^{2}}\right)``
    `` {\lambda }_{4}=410.2\,\mathrm{\,nm\,}``
    The wavelength for the transition from n = 7 to n = 2 is given by
    `` \frac{1}{{\lambda }_{5}}=R\left(\frac{1}{{2}^{2}}-\frac{1}{{7}^{2}}\right)``
    `` {\lambda }_{5}=397.0\,\mathrm{\,nm\,}``
    Thus, the wavelengths emitted by the atomic hydrogen in visible range (380-780 nm) are 5.
    Lyman series contains wavelengths ranging from 91 nm (for n2 = 2) to 121 nm (n2 =`` \infty ``).
    So, the wavelengths in the given range (50-100 nm) must lie in the Lyman series.
    The wavelength in the Lyman series can be found by
    `` \frac{1}{\lambda }=R\left(\frac{1}{{1}^{2}}-\frac{1}{{n}^{2}}\right)``
    The wavelength for the transition from n = 2 to n = 1 is given by
    `` \frac{1}{{\lambda }_{1}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)``
    `` {\lambda }_{1}=122\,\mathrm{\,nm\,}``
    The wavelength for the transition from n = 3 to n = 1 is given by
    `` \frac{1}{{\lambda }_{2}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)``
    `` {\lambda }_{2}=103\,\mathrm{\,nm\,}``
    The wavelength for the transition from n = 4 to n = 1 is given by
    `` \frac{1}{{\lambda }_{3}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{4}^{2}}\right)``
    `` {\lambda }_{3}=97.3\,\mathrm{\,nm\,}``
    The wavelength for the transition from n = 5 to n = 1 is given by
    `` \frac{1}{{\lambda }_{4}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{5}^{2}}\right)``
    `` {\lambda }_{4}=95.0\,\mathrm{\,nm\,}``
    The wavelength for the transition from n = 6 to n = 1 is given by
    `` \frac{1}{{\lambda }_{5}}=R\left(\frac{1}{{1}^{2}}-\frac{1}{{6}^{2}}\right)``
    `` {\lambda }_{5}=93.8\,\mathrm{\,nm\,}``
    So, it can be noted that the number of wavelengths lying between 50 nm to 100 nm are 3.
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