A) 27 : 5
B) 4 : 1
C) 5 : 4
D) 9 : 4
Correct Answer: D
Solution :
For Lyman \[{{\text{\bar{v}}}_{\max }}={{R}_{H}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{\infty }^{2}}} \right)={{R}_{H}}\] \[{{\text{\bar{v}}}_{\min }}={{R}_{H}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)=\frac{3}{4}{{R}_{H}}\] \[\Delta {{\text{\bar{v}}}_{Lyman}}=\frac{{{R}_{H}}}{4}\] For Balmer \[{{\text{\bar{v}}}_{\max }}={{R}_{H}}\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{\infty }^{2}}} \right)=\frac{{{R}_{H}}}{4}\] \[{{\text{\bar{v}}}_{\min }}={{R}_{H}}\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)=\frac{5}{36}{{R}_{H}}\] \[\Delta {{\text{\bar{v}}}_{Balmer}}=\frac{{{R}_{H}}}{4}-\frac{5{{R}_{H}}}{36}=\frac{4{{R}_{H}}}{36}=\frac{{{R}_{H}}}{9}\] \[\frac{\Delta {{{\text{\bar{v}}}}_{Lyman}}}{\Delta {{{\text{\bar{v}}}}_{Balmer}}}=\frac{{}^{{{R}_{H}}}/{}_{4}}{{}^{{{R}_{H}}}/{}_{9}}=\frac{9}{4}\] \[\therefore \] Ans. is
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