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Ratio of wavelength of series limit of Paschen and Brackett series for a single electronic species is :A. `(4)/(9)`B. `(12)/(7)`C. `(9)/(16)`D. `(16)/(25)`

Answer» Correct Answer - C

Correct option is C. \(\frac 9{16}\)

As we know

\(\frac 1\lambda = R_H \left(\frac 1{{n_f}^2}-\frac1{{n_i}^2}\right)\)   .....(i)

For limiting condition

\(n_i= \infty\) for both Paschen and Brackett series

\(n_f = 3\) for Paschen series

\(n_f = 4\) for Brackett series

\(\therefore \frac 1{\lambda_{Paschen}}= R_H\left(\frac1{3^2}- \frac1{\infty^2}\right)\)

\(\frac 1{\lambda_{Paschen}}=\frac {R_H}9\)     \(\because \left(\frac 1{\infty^2} \to0\right)\)

\({\lambda_{Paschen}}=\frac 9{R_H}\)      .......(ii)

\(\therefore \frac 1{\lambda_{Brackett}}= R_H\left(\frac1{4^2}- \frac1{\infty^2}\right)\)

\(\frac 1{\lambda_{Brackett}}= \frac{R_H}{16}\)      \(\because \left(\frac 1{\infty^2} \to0\right)\)

\({\lambda_{Brackett}}= \frac{16}{R_H}\)      .......(iii)

Therefore ratio of wave length of Paschen and Brackett series in limiting condition-

\(\frac{\lambda_{Paschen}}{\lambda_{Brackett}} = \frac{9/R_H}{16/R_H} = \frac 9{16}\)



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