A) \[\frac{4}{3}\]
B) \[\frac{525}{376}\]
C) \[25\]
D) \[\frac{900}{11}\]
Correct Answer: D
Solution :
Shortest wavelength comes from \[{{n}_{1}}=\infty \] to \[{{n}_{2}}=1\]and longest wavelength comes from \[{{n}_{1}}=6\] to \[{{n}_{2}}=5\] in the given case. Hence \[\frac{1}{{{\lambda }_{\min }}}=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{\infty }^{2}}} \right)=R\] \[\frac{1}{{{\lambda }_{\max }}}=R\left( \frac{1}{{{5}^{2}}}-\frac{1}{{{6}^{2}}} \right)=R\left( \frac{36-25}{25\times 36} \right)=\frac{11}{900}R\] \[\therefore \,\,\frac{{{\lambda }_{\max }}}{{{\lambda }_{\min }}}=\frac{900}{11}\]
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