NEET-XII-Physics
43: Bohr's Model and Physics of the Atom
- #7A hydrogen atom emits ultraviolet radiation of wavelength 102.5 nm. What are the quantum numbers of the states involved in the transition?Ans : Wavelength of ultraviolet radiation, `` \lambda `` = 102.5 nm = 102,5 `` \times ``10`` -9`` m
Rydberg's constant, R = 1.097`` \times `` 107 m`` -1``
Since the emitted light lies in ultraviolet range, the lines will lie in lyman series.
Lyman series is obtained when an electron jumps to the ground state (n​1 = 1) from any excited state (n​2).
Wavelength of light `` \left(\lambda \right)`` is given by
`` \frac{1}{\lambda }=R\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)``
`` \,\mathrm{\,Here\,},R=\,\mathrm{\,Rydberg\,}\,\mathrm{\,constant\,}``
`` \Rightarrow \frac{1}{102.5\times {10}^{-9}}=1.097\times {10}^{7}\left(\frac{1}{{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)``
`` \Rightarrow \frac{{10}^{9}}{102.5}=1.097\times {10}^{7}\left(1-\frac{1}{{n}_{2}^{2}}\right)``
`` \Rightarrow \frac{{10}^{2}}{102.5}=1.097\left(1-\frac{1}{{n}_{2}^{2}}\right)``
`` \Rightarrow 1-\frac{1}{{n}_{2}^{2}}=\frac{100}{102.5\times 1.097}``
`` \Rightarrow \frac{1}{{n}_{2}^{2}}=1-\frac{100}{102.5\times 1.097}``
`` \Rightarrow {n}_{2}=\sqrt{9.041}=3``
The transition will be from 1 to 3.
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