A) \[n=2\to n=1\]
B) \[n=3\to n=2\]
C) \[n=5\to n=4\]
D) \[n=8\to n=6\]
Correct Answer: D
Solution :
[d] In case of H-spectrum UV spectrum appears from \[{{n}_{1}}=1,\]to \[{{n}_{2}}=2,3....\] Here in case of\[H{{e}^{\oplus }}\]spectrum UV spectrum is from \[{{n}_{1}}=3\]3 to \[{{n}_{2}}=4\], \[{{\bar{v}}_{UV}}=R{{Z}^{2}}\left( \frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}} \right)=R\times 4\left( \frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}} \right)\] In case of H-spectrum IR appears from \[{{n}_{1}}=3.\] \[{{n}_{2}}=4,5....\] n,=4,5.... \[{{\bar{v}}_{IR}}=\] for \[{{H}_{2}}^{\oplus }=R\times {{1}^{2}}\left( \frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}} \right)\] ?..(i) \[{{v}_{IR}}\] for \[H{{e}^{\oplus }}=R\times 4\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)=R\left( \frac{4}{n_{1}^{2}}-\frac{4}{n_{2}^{2}} \right)\] ?.(ii) Comparing the coefficients of equations (i) and (ii),we get \[\left( \frac{4}{n_{1}^{2}}-\frac{1}{9} \right);\,\,{{n}_{1}}=6\] \[\left( \frac{4}{n_{2}^{2}}-\frac{1}{16} \right);\,\,{{n}_{1}}=8\] Thus transition for \[H{{e}^{\oplus }}\] in IR is \[{{n}_{1}}\to 6\] to \[{{n}_{2}}\to 8\].
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