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Chapter 6: Problem 5

A simple calculation of the energy of the helium atom supposes that each electron occupies the same hydrogenic 1 s-orbital (but with \(Z=2\) ). The electronelectron interaction is regarded as a perturbation, and calculation gives $$\int \psi_{1 s}^{2}\left(r_{1}\right)\left(\frac{e^{2}}{4 \pi \varepsilon_{0} r_{12}}\right) \psi_{1 s}^{2}\left(r_{2}\right) \mathrm{d} \tau=\frac{5}{4}\left(\frac{e^{2}}{4 \pi \varepsilon_{0} a_{0}}\right)$$ (see Example 7.2 ). Estimate (a) the binding energy of helium, (b) its first ionization energy. Hint. Use eqn 6.15 with \(E_{1}=E_{2}=E_{1 \mathrm{s}} .\) Be careful not to count the electronelectron interaction energy twice.

Short Answer

Expert verified
The binding energy of the helium atom is 74.8 eV, and its first ionization energy is 54.4 eV.

Step by step solution

01

Understand the model and break down the information

We have a model for the helium atom where both electrons act like they are in a hydrogenic atom with charge \(Z = 2\). Given is a perturbed energy term that we need to consider. It perturbs the energy level of the electron's orbital due to electron-electron interaction: \(\frac{5}{4}\left(\frac{e^{2}}{4 \pi \varepsilon_{0} a_{0}}\right)\). Here, \(e\) is the charge of the electron, \(\varepsilon_{0}\) is the permittivity of free space, and \(a_{0}\) is the Bohr radius.
02

Calculate the unperturbed energy of the helium atom

The unperturbed energy would be twice the energy of a hydrogenic 1s orbital with \(Z=2\) (since there are two electrons in helium). So, we have \(E_{\text{unperturbed}} = 2E_{1s}= 2(-13.6Z^{2} \text{ eV}) = 2(-13.6 * 2^{2}) \text{ eV} = -108.8 \text{ eV}\).
03

Calculate the perturbation of the energy due to electron-electron interaction

We use the given perturbation term which is \(\frac{5}{4}\left(\frac{e^{2}}{4 \pi \varepsilon_{0} a_{0}}\right)\) to calculate the perturbation of the energy caused by electron-electron interaction. Substituting the values of \(e, \varepsilon_{0}, a_{0}\), we find that this term equals approximately 34 eV.
04

Calculate the total energy of the Helium atom

The total energy of the Helium atom will be the sum of the unperturbed energy and the perturbation. So, we have \(E_{\text{total}} = E_{\text{unperturbed}} + \text{perturbation} = -108.8 \text{ eV} + 34 \text{ eV} = -74.8 \text{ eV}\). This means that the binding energy of the helium atom is 74.8 eV.
05

Calculate the first ionization energy

This is the energy required to remove one electron. Since one electron has its energy already calculated as -54.4 eV, the ionization energy would be the absolute value of this energy, which is 54.4 eV.

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Most popular questions from this chapter

Show group-theoretically that when a perturbation of the form \(H^{(1)}=a z\) is applied to a hydrogen atom, the 1 s-orbital is contaminated by the admixture of \(n \mathrm{p}_{z^{-}}\) orbitals. Deduce which orbitals mix into (a) \(2 p_{x}\) -orbitals, (b) \(2 \mathrm{p}_{z}\) -orbitals (c) \(3 d_{x y}\) -orbitals.

Consider the hypothetical linear \(\mathrm{H}_{3}\) molecule. The wavefunctions may be modelled by expressing them as \(\psi=c_{\Lambda} s_{A}+c_{B} s_{B}+c_{C} s_{C}\) the \(s_{i}\) denoting hydrogen 1 s-orbitals of the relevant atom. Use the Rayleigh-Ritz method to find the optimum values of the coefficients and the energies of the orbital. Make the approximations \(H_{\mathrm{ss}}=\alpha, H_{\mathrm{ss}^{\prime}}=\beta\) for neighbours but 0 for non-neighbours, \(S_{\mathrm{ss}}=1,\) and \(S_{\mathrm{ss}^{\prime}}=0\) Hint. Although the basis can be used as it stands, it leads to a \(3 \times 3\) determinant and hence to a cubic equation for the energies. A better procedure is to set up symmetry-adapted combinations, and then to use the vanishing of \(H_{i j}\) unless \(\Gamma^{(i)}=\Gamma^{(j)}\).

\(\mathrm{A}\) biradical is prepared with its two electrons in a singlet state. A magnetic field is present, and because the two electrons are in different environments their interaction with the field is \(\left(\mu_{\mathrm{B}} / \hbar\right) \mathscr{B}\left(g_{1} s_{1 z}+g_{2} s_{2 z}\right)\) with \(g_{1} \neq g_{2} .\) Evaluate the time-dependence of the probability that the electron spins will acquire a triplet configuration (that is, the probability that the \(S=1, M_{s}=0\) state will be populated). Examine the role of the energy separation \(h J\) of the singlet state and the \(M_{S}=0\) state of the triplet. Suppose \(g_{1}-g_{2}\) \(\approx 1 \times 10^{-3}\) and \(J \approx 0 ;\) how long does it take for the triplet state to emerge when \(\mathscr{B}=1.0 \mathrm{T}\) ? Hint. Use eqn 6.63 ; take \(|0,0\rangle=\left(1 / 2^{1 / 2}\right)(\alpha \beta-\beta \alpha)\) and \(|1,0\rangle=\left(1 / 2^{1 / 2}\right)(\alpha \beta+\beta \alpha) .\) See Problem 4.26 for the significance of \(\mu_{\mathrm{B}}\) and \(g\)

Calculate the first-order correction to the energy of a ground-state harmonic oscillator subject to an anharmonic potential of the form \(a x^{3}+b x^{4}\) where \(a\) and \(b\) are small (anharmonicity) constants. Consider the three cases in which the anharmonic perturbation is present (a) during bond expansion \((x \geq 0)\) and compression \((x \leq 0)\) (b) during expansion only, (c) during compression only.

A hydrogen atom in a \(2 \mathrm{s}^{1}\) configuration passes into a region where it experiences an electric field in the \(z\) -direction for a time \(\tau .\) What is its electric dipole moment during its exposure and after it emerges? Hint. Use eqn 6.62 with \(\omega_{21}=0 ;\) the dipole moment is the expectation value of \(-e z\) \\[\text { use } \int \psi_{2 \mathrm{s}} z \psi_{2 \mathrm{p}} \mathrm{d} \tau=3 a_{0}\\]
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