# The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about $9$ . The spectral series are :Option 1)Lymen and PaschenOption 2)Balmer and BrackettOption 3) Brackett and Pfund Option 4)Paschen and Pfund

$\frac{1}{\lambda }=R\left ( \frac{1}{n^{2}_{1}}-\frac{1}{n^{2}_{2}} \right )$

for shortest wavelength $n_{2}=\infty$

$\frac{1}{\lambda }=R\left ( \frac{1}{n^{2}_{1}} \right )$

For Lymen =  $n_{1}=1$

Balmer=  $n_{1}=2$

Paschen= $n_{1}=3$

Brackett=  $n_{1}=4$

Pfund=     $n_{1}=5$

$\frac{\lambda Paschen}{\lambda Lymen}=\frac{3^{2}}{1^{2}}=9$

Option 1)

Lymen and Paschen

Option 2)

Balmer and Brackett

Option 3)

Brackett and Pfund

Option 4)

Paschen and Pfund

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