Answer
(a) $5.697mm$
(b) $218eV$
Work Step by Step
(a) We can find the shortest wavelength as
$\lambda=[RZ^2(\frac{1}{n^{\prime 2}}-\frac{1}{n^2})]^{-1}$
We plug in the known values to obtain:
$\lambda=[(1.097\times 10^7)(4^2)(\frac{1}{1}-\frac{1}{\infty})]^{-1}$
$\lambda=5.697mm$
(b) We can find the required ionization energy as follows:
$|\Delta E|=(Z^2)(13.6eV)$
$|\Delta E|=(4^2)(13.6eV)$
$|\Delta E|=218eV$
