A) \[13122 \overset{{}^\circ }{\mathop{A}}\,\]
B) \[3280 \overset{{}^\circ }{\mathop{A}}\,\]
C) \[4260 \overset{{}^\circ }{\mathop{A}}\,\]
D) \[2187 \overset{{}^\circ }{\mathop{A}}\,\]
Correct Answer: C
Solution :
The wavelength of spectral line in Balmer series is given by \[\,\frac{1}{\lambda }=R\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{n}^{2}}} \right]\] For first line of Balmer series, \[\operatorname{n} = 3\] \[\Rightarrow \,\,\,\frac{1}{{{\lambda }_{1}}}=R\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right]=\frac{5R}{36}\] for second line \[\operatorname{n} = 4\] \[\Rightarrow \,\,\,\,\frac{1}{{{\lambda }_{2}}}=R\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right]=\frac{3R}{16}\] \[\therefore \,\,\,\,\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}\,\,=\,\,\frac{20}{27}\,\,\,\,\Rightarrow \,\,\,{{\lambda }_{2}}=\frac{20}{27}\times 6561=4860\,\overset{{}^\circ }{\mathop{A}}\,\]
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