NEET-XII-Physics
43: Bohr's Model and Physics of the Atom
- #13Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in Lyman series. What wavelength does this latter photon correspond to?Ans : As the second photon emitted lies in the Lyman series, the transition will be from the states having quantum numbers n = 2 to n = 1.
Wavelength of 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 is the Rydberg constant, having the value of 1.097×107 m-1.
`` \frac{1}{\lambda }=1.097\times {10}^{7}\left[\frac{1}{{\left(1\right)}^{2}}-\frac{1}{{\left(2\right)}^{2}}\right]``
`` \Rightarrow \frac{1}{\lambda }=1.097\times {10}^{7}\left[1-\frac{1}{4}\right]``
`` \Rightarrow \frac{1}{\lambda }=1.097\times \frac{3}{4}\times {10}^{7}``
`` \Rightarrow \lambda =\frac{4}{1.097\times 3\times {10}^{7}}``
`` =1.215\times {10}^{-7}``
`` =121.5\times {10}^{-9}=122\,\mathrm{\,nm\,}``
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