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Chapter 7: Problem 178

For hydrogen atoms, the wave function for the state \(n=3\) \(\ell=0, m_{\ell}=0\) is $$\psi_{300}=\frac{1}{81 \sqrt{3 \pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2}\left(27-18 \sigma+2 \sigma^{2}\right) e^{-\sigma \beta}$$ where \(\sigma=r / a_{0}\) and \(a_{0}\) is the Bohr radius $\left(5.29 \times 10^{-11} \mathrm{m}\right) .$ Calculate the position of the nodes for this wave function.

Short Answer

Expert verified
The nodes for the wave function \(\psi_{300}\) are the positions in terms of r where the probability density of finding an electron is zero. To find these positions, we set the wave function equal to zero and solve for r. We find the node positions to be: \[r_1 = \left(\frac{18 + \sqrt{216}}{4}(5.29 \times 10^{-11}\mathrm{m})\right)\] \[r_2 = \left(\frac{18 - \sqrt{216}}{4}(5.29 \times 10^{-11}\mathrm{m})\right)\]

Step by step solution

01

Identify the wave function for the given quantum numbers

We are given the wave function for the hydrogen atom in the state with quantum numbers n=3, l=0, and m_l=0: \[\psi_{300}=\frac{1}{81 \sqrt{3 \pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2}\left(27-18 \sigma+2 \sigma^{2}\right) e^{-\sigma \beta}\] where \(\sigma = r/a_0\) and \(a_0\) is the Bohr radius (\(5.29 \times 10^{-11}\) m).
02

Set the wave function equal to zero

To find the nodes, we want to find the values of r for which the wave function is equal to zero: \[\frac{1}{81 \sqrt{3 \pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2}\left(27-18 \sigma+2 \sigma^{2}\right) e^{-\sigma \beta} = 0\]
03

Identify the factors affecting the result

The exponential term \(e^{-\sigma \beta}\) is always positive and can never be equal to zero. The same goes for the constants in the expression. Therefore, the key term to focus on is the polynomial one: \[27-18 \sigma+2 \sigma^{2} = 0\] Replace \(\sigma\) with \(r/a_{0}\) to find the nodes in terms of r: \[27-18\left(\frac{r}{a_{0}}\right) + 2 \left(\frac{r}{a_{0}}\right)^2 = 0\]
04

Solve the quadratic equation for r

Solve the quadratic equation to find the roots (i.e., node positions) in terms of r: \[2\left(\frac{r}{a_{0}}\right)^2 - 18\left(\frac{r}{a_{0}}\right) + 27 = 0\] Let \(x = \frac{r}{a_{0}}\), so the equation becomes: \[2x^2 - 18x + 27 = 0\] Use the quadratic formula to solve for x: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] where a = 2, b = -18, and c = 27. \[x = \frac{18 \pm \sqrt{(-18)^2 - 4(2)(27)}}{2(2)}\] Simplify and solve for x: \[x = \frac{18 \pm \sqrt{216}}{4}\] This gives us two values for \(x\): \[x_1 = \frac{18 + \sqrt{216}}{4}\] \[x_2 = \frac{18 - \sqrt{216}}{4}\]
05

Convert the x values back to r-values

Recall that \(x = \frac{r}{a_{0}}\). Multiply both sides by \(a_{0}\) to obtain the node positions, \(r_1\) and \(r_2\): \[r_1 = x_1a_0 = \left(\frac{18 + \sqrt{216}}{4}\right)(5.29 \times 10^{-11}\mathrm{m})\] \[r_2 = x_2a_0 = \left(\frac{18 - \sqrt{216}}{4}\right)(5.29 \times 10^{-11}\mathrm{m})\] Now you have found the position of the nodes for this wave function.

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Most popular questions from this chapter

It takes 476 kJ to remove 1 mole of electrons from the atoms at the surface of a solid metal. What is the maximum wavelength of light that can remove a single electron from an atom at the surface of this solid metal?

Answer the following questions based on the given electron configurations, and identify the elements. a. Arrange these atoms in order of increasing size: $[\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{6} ;[\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{1} ;[\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{3}$ b. Arrange these atoms in order of decreasing first ionization energy: [Ne $3 s^{2} 3 p^{5} ;[\operatorname{Ar}] 4 s^{2} 3 d^{10} 4 p^{3} ;[\operatorname{Ar}] 4 s^{2} 3 d^{10} 4 p^{5}$

Give the maximum number of electrons in an atom that can have these quantum numbers: a. \(n=0, \ell=0, m_{\ell}=0\) b. \(n=2, \ell=1, m_{\ell}=-1, m_{s}=-\frac{1}{2}\) c. \(n=3, m_{s}=+\frac{1}{2}\) d. \(n=2, \ell=2\) e. \(n=1, \ell=0, m_{\ell}=0\)

Predict the atomic number of the next alkali metal after francium and give its ground-state electron configuration.

The work function of an element is the energy required to remove an electron from the surface of the solid element. The work function for lithium is 279.7 kJ/mol (that is, it takes 279.7 kJ of energy to remove one mole of electrons from one mole of Li atoms on the surface of Li metal). What is the maximum wavelength of light that can remove an electron from an atom on the surface of lithium metal?
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