A) Paschen and P fund
B) Lyman and Paschen
C) Brackett and Piund
D) Balmer and Brackett
Correct Answer: B
Solution :
\[\frac{\frac{1}{{{\lambda }_{2}}}={{R}_{H}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right){{Z}^{2}}}{\frac{1}{{{\lambda }_{1}}}={{R}_{H}}\left( \frac{1}{m_{1}^{2}}-\frac{1}{m_{2}^{2}} \right){{Z}^{2}}}\]as for shortest wavelengths both \[{{n}_{2}}\]and \[{{m}_{2}}\]are\[\infty \] \[\therefore \]\[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{9}{1}=\frac{m_{1}^{2}}{n_{1}^{2}}\] Now if \[{{m}_{1}}=3\And {{n}_{1}}=1\]it will justify the statement hence Lyman and Paschen is correct.
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