If Rydberg's constant is $\mathrm{R}$, the longest wavelength of radiation in Paschen series will be $\frac{\alpha }{7R}$, where $\alpha =$______.

In the Paschen radiation series, the longest wavelength corresponds to the transition between $n=3$ and $n=4$.

The formula for the wavelength in the given transition can be written as, 

$\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) ...\left(1\right)$

From equation (1), it follows that

$$\begin{gathered} \frac{1}{\lambda }=R{Z}^2\left(\frac{{\displaystyle 1}}{{\displaystyle 3^2}}-\frac{{\displaystyle 1}}{{\displaystyle 4^2}}\right) \\ =R{Z}^2\left(\frac{{\displaystyle 1}}{{\displaystyle 9}}-\frac{{\displaystyle 1}}{{\displaystyle 16}}\right) \\ =\frac{7R{Z}^2}{9\times 16} \\ \Rightarrow \lambda =\frac{144}{7\mathrm{R}} \end{gathered}$$

 Hence, $\alpha =144$.