Question

a hydrogen - like ion is an ion containing only one electron. the energies of the electron in a hydrogen - like ion are given by

$e_n=-(2180\times10^{-18}j)(\frac{z^2}{n^2})$

where $n$ is the principal quantum number, and $z$ is the atomic number of the element. calculate the ionization energy, in $\frac{kj}{mol}$, of the $be^{3 - }$ ion. be sure your answer has the correct number of significant digits.

note: reference the fundamental constants table for additional information.

$\frac{kj}{mol}$

Explanation:

Step1: Identify \(Z\) and \(n\) values

For \(Be^{3 -}\), the atomic number \(Z = 4\). Ionization energy is the energy required to remove an electron from \(n=\infty\) (where \(E = 0\)) to \(n = 1\) (ground - state). So, \(n = 1\).

Step2: Calculate energy for one ion

Use the formula \(E_n=-(2.18\times 10^{-18}\text{ J})\frac{Z^{2}}{n^{2}}\).
Substitute \(Z = 4\) and \(n = 1\) into the formula:
\(E_1=-(2.18\times 10^{-18}\text{ J})\frac{4^{2}}{1^{2}}=-(2.18\times 10^{-18}\text{ J})\times16=- 3.488\times 10^{-17}\text{ J}\)
The ionization energy for one ion \(\Delta E=E_{\infty}-E_1\). Since \(E_{\infty} = 0\), \(\Delta E=3.488\times 10^{-17}\text{ J}\) per ion.

Step3: Convert to energy per mole

Use Avogadro's number \(N_A = 6.022\times 10^{23}\text{ mol}^{-1}\).
\(\Delta E_{\text{mol}}=(3.488\times 10^{-17}\text{ J})\times(6.022\times 10^{23}\text{ mol}^{-1})\)
\(\Delta E_{\text{mol}}=3.488\times6.022\times 10^{6}\text{ J/mol}\)
\(\Delta E_{\text{mol}} = 2.10\times 10^{7}\text{ J/mol}\)
Convert to \(\text{kJ/mol}\): \(\Delta E_{\text{mol}}=\frac{2.10\times 10^{7}\text{ J/mol}}{10^{3}} = 2.10\times 10^{4}\text{ kJ/mol}\)

Answer:

\(2.10\times 10^{4}\text{ kJ/mol}\)