A) \[X\frac{108}{80}\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) \[X\frac{80}{108}\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) \[\frac{1}{X}\frac{80}{108}\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) \[\frac{1}{X}\frac{108}{80}\overset{\text{o}}{\mathop{\text{A}}}\,\]
Correct Answer: B
Solution :
[b] \[{{H}_{\alpha }}\]line of Balmer series means first line of Balmer series. \[{{n}_{1}}=2,{{n}_{2}}=3\] \[\overline{v}=\frac{1}{{{\lambda }_{\alpha }}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)=\frac{5R}{36}\] \[\therefore {{\lambda }_{\alpha }}=\frac{36}{5R}=X\] \[{{H}_{\beta }}\]line of Balmer series means, second line of Balmer series, \[{{n}_{1}}=2,{{n}_{2}}=4\] \[\overline{v}=\frac{1}{{{\lambda }_{\beta }}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)=\frac{3R}{16}\] \[\therefore {{\lambda }_{\beta }}=\frac{16}{3R}=X\] when \[\frac{36}{3R}=X\] then \[\frac{16}{3R}=\frac{X\times 5R\times 16}{36\times 3R}=\frac{80X}{108}\overset{{}^\circ }{\mathop{A}}\,\]
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