A) 5 : 9
B) 5 : 36
C) 1 : 4
D) 3 : 4
Correct Answer: A
Solution :
For H-atom \[\frac{1}{\lambda }=R\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] For 1st line of Balmer series ie, \[{{n}_{1}}=2,\]and \[{{n}_{2}}=3\] wavelength is maximum \[\Rightarrow \] \[\frac{1}{{{\lambda }_{\max }}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)\] \[=\frac{5R}{36}\] or \[{{\lambda }_{\max }}=\frac{36}{5R}\] For minimum wavelength \[{{n}_{1}}=2,\] \[{{n}_{2}}=\infty \] \[\therefore \] \[\frac{1}{{{\lambda }_{\min }}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{\infty }^{2}}} \right)\] \[=\frac{R}{4}\] \[\Rightarrow \] \[{{\lambda }_{\min }}=\frac{4}{R}\] Therefore, \[\frac{{{\lambda }_{\min }}}{{{\lambda }_{\max }}}=\frac{4}{R}\times \frac{5R}{36}=\frac{5}{9}\] SO, \[{{\lambda }_{\min }}:{{\lambda }_{\max }}=5:9\]
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