Explanation

(b) The series in U-V region is Lyman series. Longest wavelength corresponds to, minimum energy which occurs in transition from $n$=2 to $n$=1.
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 122 = \frac{\frac{1}{R}}{\left( \frac{1}{1^{2}} - \frac{1}{2^{2}} \right)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
The smallest wavelength in the infrared region corresponds to maximum energy of Paschen series.
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda = \frac{\frac{1}{R}}{\left( \frac{1}{3^{2}} - \frac{1}{\ \infty} \right)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Solving Eqs.(i) and (ii) , we get
$\lambda = 823.5\ nm$