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 Paschen

  • Option 2)

    Balmer and Brackett

  • Option 3)

     Brackett and Pfund 

  • Option 4)

    Paschen and Pfund

Answers (1)

\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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