A) \[13122\ {AA}\]
B) \[3280\ {AA}\]
C) \[4860\ {AA}\]
D) \[2187\ {AA}\]
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, n = 3 Þ \[\frac{1}{{{\lambda }_{1}}}=R\,\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right]=\frac{5R}{36}\]; For second line n = 4. Þ \[\frac{1}{{{\lambda }_{2}}}=R\,\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right]=\frac{3R}{16}\] \ \[\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{20}{27}\Rightarrow {{\lambda }_{1}}=\frac{20}{27}\times 6561=4860\,{\AA}\]
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