A) 802 nm
B) 823 nm
C) 1882 nm
D) 1648 nm
Correct Answer: B
Solution :
[b]: The smallest frequency and largest wavelength in ultraviolet region will be for transition from \[{{n}_{2}}=2\]to \[{{n}_{1}}=1\] \[\therefore \]\[\frac{1}{{{\lambda }_{\max }}}=R\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] \[\Rightarrow \]\[\frac{1}{{{\lambda }_{\max }}}=R\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right]=R\left[ 1-\frac{1}{4} \right]=\frac{3R}{4}\] ?(i) The highest frequency and smallest wavelength for infrared region will be for transition from\[{{n}_{2}}=\infty \]to\[{{n}_{1}}=3\] \[\therefore \]\[\frac{1}{{{\lambda }_{\min }}}=R\left( \frac{1}{{{3}^{2}}}-\frac{1}{\infty } \right)=\frac{R}{9}\]?(ii) From (i) and (ii), we get \[\frac{{{\lambda }_{\min }}}{{{\lambda }_{\max }}}=\frac{3R}{4}\times \frac{9}{R}=\frac{27}{4}\] \[{{\lambda }_{\min }}=\frac{27}{4}\times {{\lambda }_{\max }}\] \[=\frac{27}{4}\times 122nm=823.5nm\]
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