NEET-XII-Physics
43: Bohr's Model and Physics of the Atom
- #9A group of hydrogen atoms are prepared in n = 4 states. List the wavelength that are emitted as the atoms make transitions and return to n = 2 states.Ans : There will be three wavelengths.
(i) For the transition from (n = 4) to (n = 3) state
(ii) For the transition from (n = 3) to (n = 2) state
(iii) For the transition from (n = 4) to (n = 2) state
Let (`` \lambda ``1) be the wavelength when the atom makes transition from (n = 4) state to (n = 2) state.​
Here,
n1 = 2
n2 = 4
Now, the wavelength (`` \lambda ``1)​ will be
`` \frac{1}{{\lambda }_{1}}=R\left(\frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}}\right)``
`` ``
`` ``
`` R=1.097\times {10}^{7}{\,\mathrm{\,m\,}}^{-1}``
`` ``
`` \frac{1}{{\lambda }_{1}}=1.097\times {10}^{7}\times \left(\frac{1}{4}-\frac{1}{16}\right)``
`` \Rightarrow \frac{1}{{\lambda }_{1}}=1.097\times {10}^{7}\left(\frac{4-1}{16}\right)``
`` \Rightarrow \frac{1}{{\lambda }_{1}}=\frac{1.097\times {10}^{7}\times 3}{16}``
`` \Rightarrow {\lambda }_{1}=\frac{16\times {10}^{-7}}{3\times 1.097}``
`` =4.8617\times {10}^{-7}``
`` =486.1\times {10}^{-9}``
`` =487\,\mathrm{\,nm\,}``
When an atom makes transition from (n = 4) to (n = 3), the wavelength (`` \lambda ``2) is given by
`` \,\mathrm{\,Here\,}\,\mathrm{\,again\,},``
`` {n}_{1}=3``
`` {n}_{2}=4``
`` \frac{1}{{\lambda }_{2}}=1.097\times {10}^{7}\left(\frac{1}{9}-\frac{1}{16}\right)``
`` \Rightarrow \frac{1}{{\lambda }_{2}}=1.097\times {10}^{7}\left(\frac{16-9}{144}\right)``
`` \Rightarrow \frac{1}{{\lambda }_{2}}=\frac{1.097\times {10}^{7}\times 7}{144}``
`` \Rightarrow {\lambda }_{2}=\frac{144}{1.097\times {10}^{7}\times 7}``
`` =1875\,\mathrm{\,nm\,}``
Similarly, wavelength (`` \lambda ``3) for the transition from (n = 3) to (n = 2) is given by
When the transition is n1 = 2 to n2 = 3:
`` \frac{1}{{\lambda }_{3}}=1.097\times {10}^{7}\left(\frac{1}{4}-\frac{1}{9}\right)``
`` \Rightarrow \frac{1}{{\lambda }_{3}}=1.097\times {10}^{7}\left(\frac{9-4}{36}\right)``
`` \Rightarrow \frac{1}{{\lambda }_{3}}=\frac{1.097\times {10}^{7}\times 5}{36}``
`` \Rightarrow {\lambda }_{3}=\frac{36\times {10}^{-7}}{1.097\times 5}=656\,\mathrm{\,nm\,}``
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