Physics - Modern Physics Question with Solution | TestHub

The series in $U-V$ region is Lymen series. Longest wavelength corresponds to minimum energy which occurs in transition from $n=2$ to $n=1$. $$\therefore 122=\frac{\frac{1}{R}}{\left(\frac{1}{1^2}-\frac{1}{2^2}\right)}$$ 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). $$\lambda =823.5\mathrm{nm}$$