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, \[{{n}_{1}}=1,\,{{n}_{2}}=2\] \[\frac{1}{{{\lambda }_{L}}}=R\left( \frac{1}{1}-\frac{1}{{{2}^{2}}} \right)=\frac{3R}{4}\] ?(i) For \[{{H}_{\alpha }}\]line, \[{{n}_{1}}=2,\,{{n}_{3}}=3\] \[\frac{1}{{{\lambda }_{{{H}_{\alpha }}}}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)=\frac{5R}{36}\] ?(ii) From Eqs. (i) and (ii), we get \[\frac{{{\lambda }_{{{H}_{a}}}}}{{{\lambda }_{L}}}=\frac{3}{4}\times \frac{36}{5}=\frac{27}{5}\] \[\Rightarrow \] \[{{\lambda }_{{{H}_{\alpha }}}}=\frac{27}{5}\times 1215=6561\overset{\text{o}}{\mathop{\text{A}}}\,\]
You need to login to perform this action.
You will be redirected in
3 sec
