To find the speed of the electron in the third orbit of $\mathrm{He}^+$, we use the Bohr model of the atom.$\\$In the Bohr model, the speed of an electron in the $n$-th orbit is given by:$\\ v_n = \frac{Z e^2}{2 \varepsilon_0 h} \cdot \frac{1}{n} \\$Where:$\\$• $Z$ is the atomic number (for $\mathrm{He}^+, Z = 2) \\$• $e$ is the charge of the electron $(1.6 \times 10^{-19} \, \mathrm{C}) \\$• $\varepsilon_0$ is the permittivity of free space $(8.85 \times 10^{-12} \, \mathrm{C^2/N \cdot m^2}) \\$• $h$ is Planck's constant $(6.63 \times 10^{-34} \, \mathrm{Js}) \\$• $n$ is the orbit number (for the third orbit, $n = 3) \\$Substitute the known values into the formula:$\\ v_3 = \frac{2 \cdot (1.6 \times 10^{-19})^2}{2 \cdot 8.85 \times 10^{-12} \cdot 6.63 \times 10^{-34}} \cdot \frac{1}{3} \\$Simplify the expression:$\\ v_3 = \frac{2 \cdot 2.56 \times 10^{-38}}{2 \cdot 8.85 \times 10^{-12} \cdot 6.63 \times 10^{-34}} \cdot \frac{1}{3} \\ v_3 = \frac{5.12 \times 10^{-38}}{1.173 \times 10^{-45}} \cdot \frac{1}{3} \\ v_3 = \frac{5.12}{1.173} \times 10^7 \cdot \frac{1}{3} \\ v_3 \approx 4.36 \times 10^7 \cdot \frac{1}{3} \\ v_3 \approx 1.46 \times 10^6 \, \mathrm{ms}^{-1} \\$Therefore, the speed of the electron in the third orbit of $\mathrm{He}^+$ is $1.46 \times 10^6 \, \mathrm{ms}^{-1}. \\$This corresponds to Option 1.