The velocity of the electrons liberated by electromagnetic radiation of wavelength $\mathrm{\lambda }=18.0 \mathrm{nm}$, from stationary ${\mathrm{He}}^{+}$ ions in the ground state, is

Step I : Determine binding energy of electron :

As we know    $E=\frac{-13\text{.}6{\text{Z}}^2}{n^2}=\frac{-13\text{.}6{\left(2\right)}^2}{{\left(1\right)}^2}\left(\text{For ground state}\right)$

                           = – 54.4 eV

∴ The binding energy of electron is  E_b = – E = 54.4 eV

Step II : Determine kinetic energy of electron.

∴               $T_e=\frac{hc}{\lambda }-E_b$

                      $=\frac{1242n\text{meV}}{18nm}-54\text{.}4\text{eV}$

                       = (69 – 54.4) eV = 14.6 eV

                        = 23.36 × 10^–19 joule

∴                $T_e=\frac{1}{2}{\text{mv}}^2$

                      $v=\sqrt{\frac{2Te}{m}}$

                        $=\sqrt{\frac{2\times 23\text{.}36\times 10^{-19}}{9\text{.}1\times 10^{-31}}}\text{m/s}=2\text{.}3\times 10^6\text{m/s}$