Question

A \(\mathrm{He}^{+}\) ion consists of a nucleus (containing two protons and two neutrons) and a single electron. Find the Bohr radius for this system.

Short answer

Answer: The Bohr radius for a Helium ion (He+) is approximately \(2.646 \times 10^{-11}\,\text{m}\).

Step-by-step solution

Recall the Bohr radius formula for hydrogen-like ions

The formula for the Bohr radius in a hydrogen-like ion is given by: \(r_{n} = \frac{4 \pi \varepsilon_{0} \hbar^{2} n^{2}}{Z q_{e}^{2} m_{e}}\) where \(r_{n}\) is the Bohr radius for the nth energy level, \(\varepsilon_{0}\) is the vacuum permittivity, \(\hbar\) is the reduced Planck's constant, \(n\) is the principal quantum number, \(Z\) is the atomic number, \(q_{e}\) is the elementary charge, and \(m_{e}\) is the mass of the electron. For the case of He+ ion, we have the following values: - \(n = 1\) (we are finding the radius for the lowest energy level) - \(Z = 2\) (there are two protons in the nucleus of the He+ ion) - Vacuum permittivity (\(\varepsilon_{0} = 8.854 \times 10^{-12}\,\text{C}^2/\text{N}\cdot\text{m}^2\)) - Reduced Planck's constant (\(\hbar = 1.054 \times 10^{-34}\,\text{Js}\)) - Elementary charge (\(q_{e} = 1.602 \times 10^{-19}\,\text{C}\)) - Electron mass (\(m_{e} = 9.109 \times 10^{-31}\,\text{kg}\)).

Calculate the Bohr radius for He+ ion

Now, we can plug the given values into the formula to find the Bohr radius for the He+ ion: \(r_{1} = \frac{4 \pi (8.854 \times 10^{-12}) (1.054 \times 10^{-34})^{2} (1)^{2}}{(2) (1.602 \times 10^{-19})^{2} (9.109 \times 10^{-31})}\) \(r_{1} \approx 2.646 \times 10^{-11}\,\text{m}\) So, the Bohr radius for the He+ ion is approximately \(2.646 \times 10^{-11}\,\text{m}\).