A) \[\frac{x}{4}c{{m}^{-1}}\]
B) \[x\,c{{m}^{-1}}\]
C) \[4x\,c{{m}^{-1}}\]
D) \[16x\,c{{m}^{-1}}\]
Correct Answer: C
Solution :
\[\overline{v}\] (wave number) \[{{\overline{R}}_{H}}{{Z}^{2}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] \[{{\overline{v}}_{1}}(H{{e}^{+}},Z=2)={{\overline{R}}_{H}}{{(2)}^{2}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] \[{{\overline{v}}_{2}}(B{{e}^{3+}},Z=4)={{\overline{R}}_{H}}={{(4)}^{2}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] \[\therefore \] \[\frac{{{v}_{2}}}{{{v}_{1}}}=\frac{{{(4)}^{2}}}{{{(2)}^{2}}}=\frac{16}{4}=4\] \[\therefore \] \[{{\overline{v}}_{2}}=4{{\overline{v}}_{1}}\] \[{{\overline{v}}_{2}}=4\times c{{m}^{-1}}\]
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