A) \[6561\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) \[5464\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) \[800\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) \[4840\overset{\text{o}}{\mathop{\text{A}}}\,\]
Correct Answer: A
Solution :
Key Idea: For \[{{H}_{\alpha }}\] line\[{{n}_{1}}=2,\,\,{{n}_{2}}=3\]. From the relation \[\frac{1}{\lambda }=R\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] For Lyman series, \[\frac{1}{{{\lambda }_{L}}}=R\left( \frac{1}{1}-\frac{1}{{{2}^{2}}} \right)=\frac{3R}{4}\] ? (i) For \[{{H}_{\alpha }}\] line,\[{{n}_{1}}=2,\,\,{{n}_{2}}=3\] \[\frac{1}{{{\lambda }_{{{H}_{\alpha }}}}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)=\frac{5R}{36}\] ? (ii) From Eqs. (i) and (ii), we get \[\frac{{{\lambda }_{{{H}_{\alpha }}}}}{{{\lambda }_{L}}}=\frac{3}{4}\times \frac{36}{5}=\frac{27}{5}\] \[\Rightarrow \] \[{{\lambda }_{{{H}_{\alpha }}}}=\frac{27}{5}\times 1215=6561{\AA}\]
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