According to Bohr's model,the energy of an electron is given by $E = -\frac{2.176 \times 10^{-18}}{n^2} \, J \, atom^{-1}$. Calculate the minimum energy required to remove an electron from the $3^{rd}$ orbit of a $He^{+}$ ion,and also calculate the corresponding wavelength of the photon.

The energy of an electron in the $n^{th}$ Bohr orbit for a hydrogen-like species is given by $E_n = -\frac{2.176 \times 10^{-18} \times Z^2}{n^2} \, J$.
For $He^{+}$ ion,the atomic number $Z = 2$. For the $3^{rd}$ orbit,$n = 3$.
Energy of the electron in the $3^{rd}$ orbit: $E_3 = -\frac{2.176 \times 10^{-18} \times 2^2}{3^2} = -\frac{2.176 \times 10^{-18} \times 4}{9} \, J \approx -9.67 \times 10^{-19} \, J$.
The energy required to remove the electron (ionization energy) is $\Delta E = E_{\infty} - E_3 = 0 - (-9.67 \times 10^{-19} \, J) = 9.67 \times 10^{-19} \, J$.
The wavelength $\lambda$ is calculated using $\lambda = \frac{hc}{\Delta E}$.
$\lambda = \frac{6.626 \times 10^{-34} \, J \cdot s \times 3 \times 10^8 \, m/s}{9.67 \times 10^{-19} \, J} \approx 2.055 \times 10^{-7} \, m$.