Question

Find the Bohr radius of doubly ionized lithium (Li \(^{2+}\) ).

Short answer

Answer: The Bohr radius of doubly ionized lithium (Li \(^{2+}\)) is approximately \(5.3\times 10^{-11}\ \text{m}\).

Step-by-step solution

Write down the formula for the Bohr radius for hydrogen-like atoms

The Bohr radius for a hydrogen-like atom/ion is given by the formula: \(r_{n,Z}=\frac{n^{2}h^{2}\varepsilon _{0}}{\pi m_{e}Ze^{2}}\), where \(r_{n,Z}\) is the Bohr radius, \(n\) is the principal quantum number (n=1 for the ground state), \(h\) is the Planck constant, \(\varepsilon _{0}\) is the vacuum permittivity, \(\pi\) is the well-known constant, \(m_{e}\) is the electron's mass, \(Z\) is the atomic number of the nucleus (3 for Li), and \(e\) is the elementary charge.

Use the values of constants

Now we need to plug the values of the constants into the formula. The constants are: \(h=6.626\times 10^{-34}\ \text{Js}\), \(\varepsilon _{0}=8.854\times 10^{-12}\ \text{C}^{2}\text{N}^{-1}\text{m}^{-2}\), \(m_{e}=9.109\times 10^{-31}\ \text{kg}\), \(e=1.602\times 10^{-19}\ \text{C}\).

Calculate the Bohr radius of Li \(^{2+}\)

Plug the values of the constants and the values of \(n=1\) and \(Z=3\) for Li \(^{2+}\) into the Bohr radius formula: \(r_{=1,3}=\frac{(1)^{2}(6.626\times 10^{-34}\ \text{Js})^{2}(8.854\times 10^{-12}\ \text{C}^{2}\text{N}^{-1}\text{m}^{-2})}{\pi (9.109\times 10^{-31}\ \text{kg})(3)(1.602\times 10^{-19}\ \text{C})^{2}}\). Now we can compute the Bohr radius: \(r_{=1,3}\approx 5.3\times 10^{-11}\ \text{m}\). The Bohr radius of doubly ionized lithium (Li \(^{2+}\) ) is approximately \(5.3\times 10^{-11}\ \text{m}\).