A) blue
B) yellow
C) red
D) violet
Correct Answer: A
Solution :
When an atom comes down from some higher energy level to the second energy \[(n=2),\] then the lines of spectrum are obtained in visible part and give the Balmer series. \[\frac{1}{\lambda }=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{n}^{2}}} \right),\,\,\,\,\,\,\,\,n=3,4,5...\] For second line \[n=4\] \[\therefore \] \[\frac{1}{\lambda }=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)=\frac{3R}{16}\] \[\Rightarrow \] \[\lambda =\frac{16}{3R}\] \[R=1.097\times {{10}^{7}}{{m}^{-1}}\] \[\lambda =\frac{16}{3\times 1.097\times {{10}^{7}}}\] \[=4860\times {{10}^{-10}}m\] \[\Rightarrow \] \[\lambda =4860\overset{\text{o}}{\mathop{\text{A}}}\,\] which corresponds to colour blue.
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