The largest wavelength in the ultraviolet region of the hydrogen...

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