The formula for calculating the energies of an electron in a hydrogen-like ion is \( E_{n}=-\left(2.18 \times 10^{-18} \mathrm{~J}\right) Z^{2}\left(\frac{1}{n^{2}}\right) \) This equation cannot be applied to many-electron atoms. One way to modify it for the more complex atoms is to replace \(Z\) with \((Z-\sigma)\), in which \(Z\) is the atomic number and \(\sigma\) is a positive dimensionless quantity called the shielding constant. Consider the helium atom as an example. The physical significance of \(\sigma\) is that it represents the extent of shielding that the two 1 s electrons exert on each other. Thus, the quantity \((Z-\sigma)\) is appropriately called the "effective nuclear charge." Calculate the value of \(\sigma\) if the first ionization energy of helium is \(3.94 \times\) \(10^{-18} \mathrm{~J}\) per atom. (Ignore the minus sign in the given equation in your calculation.).

Short Answer

Expert verified
To determine \(\sigma\), solve the equation \(3.94 \times 10^{-18} = (2.18 \times 10^{-18}(2-\sigma)^2)\) which yields \(\sigma = 0.51\). Therefore, the shielding constant for helium is 0.51.

Step by step solution

01

Write down the given parameters

First, write down the given parameters. These include the ionization energy of helium \(E=3.94 \times 10^{-18}\) J, the atomic number of helium \(Z=2\), and the equation \(E_{n}=-\left(2.18 \times 10^{-18}\) J \right) Z^{2}\left(\frac{1}{n^{2}}\right)\). Remember, we're ignoring the negative sign in the given equation as per the instructions in the question.
02

Replace Z in the formula with (Z - sigma)

The next step involves adjusting the equation such that it fits for the helium atom, not a hydrogen-like ion. This is done by replacing \(Z\) in the formula with \((Z - \sigma)\), where \(\sigma\) represents the shielding effect in helium. Therefore, the equation becomes: \(E_{n}= \left(2.18 \times 10^{-18}\) J \right) (Z - \sigma)^{2}\left(\frac{1}{n^{2}}\right)\).
03

Solve the equation for sigma

Now we plug our given values into our adjusted equation. With a first ionization energy, we consider the energy level \(n=1\). This gives: \(3.94 \times 10^{-18}\) J = \(\left(2.18 \times 10^{-18}\) J \right) (2 - \sigma)^{2}\). Solving this equation for \(\sigma\) provides the solution.

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