A) \[2000\overset{o}{\mathop{\text{A}}}\,,3000\overset{o}{\mathop{\text{A}}}\,\]
B) \[1575\overset{o}{\mathop{\text{A}}}\,,2960\overset{o}{\mathop{\text{A}}}\,\]
C) \[6529\overset{o}{\mathop{\text{A}}}\,,4280\overset{o}{\mathop{\text{A}}}\,\]
D) \[6552\overset{o}{\mathop{\text{A}}}\,,4863\overset{o}{\mathop{\text{A}}}\,\]
Correct Answer: D
Solution :
The wavelength of the lines in Balmer series is represented by, \[\frac{1}{\lambda }={{R}_{H}}\left[ \frac{1}{{{2}^{2}}}-\frac{1}{n_{2}^{2}} \right]\] For first wavelength \[\frac{1}{{{\lambda }_{1}}}=1.097\times {{10}^{7}}\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right]\] \[=1.524\times {{10}^{6}}m\] Or \[{{\lambda }_{1}}=\frac{1}{1.524\times {{10}^{6}}}=6.562\times {{10}^{-7}}\] Or \[{{\lambda }_{1}}=6562\overset{o}{\mathop{\text{A}}}\,\] For second wavelength \[\frac{1}{{{\lambda }_{1}}}=1.097\times {{10}^{-7}}\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right]\] \[=2.056\times {{10}^{6}}\] Or \[{{\lambda }_{2}}=\frac{1}{2.056\times {{10}^{6}}}\] Or \[{{\lambda }_{2}}=4.863-7\times 10\times {{10}^{-10}}\] Or \[{{\lambda }_{2}}=4863\overset{o}{\mathop{\text{A}}}\,\]
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