Polymers — Chemistry STD 12 Science — Question

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{Z}^{-}$$\xrightarrow[{{\text{Sublimation}}}]{{{k_s}}} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Z}+\mathrm{Br}^{-}$

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{Z}^{-}$$\xrightarrow[{{\text{elimination}}}]{{{k_e}}}\mathrm{CH}_{3} \mathrm{CH}= \mathrm{CH}_{2} +\mathrm{HZ}+\mathrm{Br}^{-}$

where 

$\mathrm{Z}^{-}=\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}(\mathrm{A})$ or $\begin{array}{*{20}{c}}
  {\,C{H_3}} \\ 
  {|\,\,\,\,\,} \\ 
  {C{H_3} - C - {O^ - }(B)} \\ 
  {|\,\,\,\,} \\ 
  {\,\,C{H_3}} 
\end{array}$

$\mathrm{k}_{\mathrm{s}}$ and $\mathrm{k}_{\mathrm{e}},$ are $,$ respectively, the rate constants for the substitution and elimination, and $\mu=\frac{\mathrm{k}_{\mathrm{s}}}{\mathrm{k}_{\mathrm{e}}},$ the correct options is