A single electron orbits around a stationary nucleus of charge +Ze where Z is a constant and e is the magnitude of electronic charge. It requires 47.2 eV to excite the electron from the second orbit to third orbit. The wavelength of electromagnetic radiation required to remove the electron from first Bohr orbit to infinity.

The energy required to excite the electron from $n_1$to $n_2$ orbit revolving around the nucleus with charge +Ze is give by 

   $\begin{array}{l}{E}_{n2}-E_{n1}=\frac{m{e}^4}{8{\epsilon }_0^2{h}^2}\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]Z^2\\ or, E_{n2}-E_{n1}=Z^2\times \left(13.6\right)\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]\end{array}$

Since 47.2 eV energy is required to excite the electro form  $n_1=2 to n_2=3$ orbit, we have $47.2=Z^2\times 13.6\left[\frac{1}{2^2}-\frac{1}{3^2}\right]$

$\begin{array}{l}\Rightarrow Z^2=\frac{47.2\times 13.6}{13.6\times 5}=24.988=25\\ \Rightarrow Z=5\end{array}$

The energy required to remove the electron from the first Bohr orbit to infinity $\left(\infty \right)$ is given by $E_{\infty }-E_1=13.6\times Z^2\left[\frac{1}{1^2}-\frac{1}{{\infty }^2}\right]=13.6\times 25 eV =340 eV$

In order to calculate the wavelength of radiation, we use Bohr's frequency relation 

$\begin{array}{l}hf=\frac{hc}{\lambda }=13.6\times 25\times \left(1.6\times {10}^{-19}\right)J\\ or \lambda =\frac{\left(6.6\times {10}^{-34}\right)\times 3\times {10}^8}{13.6\times 25\times \left(1.6\times {10}^{-19}\right)}=34.4\AA \end{array}$