A) \[\text{9780}\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) \[\text{48660}\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) \[\text{8857}\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) \[\text{4429}\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
Correct Answer: B
Solution :
Key Idea: For Balmer series an atom comes down from some higher energy level to second energy level. The wavelength of 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, \[n=3\] \[\frac{1}{{{\lambda }_{1}}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)=\frac{5\,R}{36}\] ?. (i) For second line, \[n=4\] \[\therefore \] \[\frac{1}{{{\lambda }_{2}}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)=\frac{3\,R}{16}\] ?. (ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{5\,R}{36}\times \frac{16}{3\,R}\] \[\Rightarrow \] \[{{\lambda }_{2}}=\frac{20}{27}{{\lambda }_{1}}\] \[=\frac{20}{27}\times 6561=4860\,\overset{o}{\mathop{A}}\,\]
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