Answer
(a) $121.5nm$
(b) $820.4nm$
Work Step by Step
(a) We can find the longest wavelength in Layman series as
$\lambda_2=\frac{1}{R(\frac{1}{1^2}-\frac{1}{2^2})}$
$\lambda_2=121.5nm$
(b) The shortest wavelength in Paschen series can be determined as
$\lambda_{\infty}=\frac{1}{R(\frac{1}{3^2}-0)}$
$\lambda_{\infty}=820.4nm$
