A) \[949.8\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) \[972.6\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) \[1215.8\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) \[1025.8\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
Correct Answer: C
Solution :
\[\frac{1}{\lambda }{{R}_{H}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] For longest wavelength in Lyman series, \[{{n}_{1}}=1,\text{ }{{n}_{2}}=2\] \[\frac{1}{\lambda }=1.09678\times {{10}^{7}}\left[ \frac{1}{{{(1)}^{2}}}-\frac{1}{{{(2)}^{2}}} \right]\] \[\frac{1}{\lambda }=\frac{1.09678\times {{10}^{7}}\times 3}{4}\] \[\therefore \] \[\lambda =1.2156\times {{10}^{-7}}m=1215.6{AA}\]
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