The 5s5s electron in rubidium (Rb) sees an effective charge of 2.771e2.771e. Calculate the ionization energy of this electron.

Ch 41: Quantum Mechanics II: Atomic Structure

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 41: Quantum Mechanics II: Atomic Structure

Problem 31

Chapter 41, Problem 31

The $5 s$ electron in rubidium (Rb) sees an effective charge of $2.771 e$ . Calculate the ionization energy of this electron.

Verified step by step guidance

Step 1: Understand the concept of ionization energy. Ionization energy is the energy required to remove an electron from an atom in its ground state. For this problem, we will use the effective nuclear charge (Z_eff) and the Bohr model to calculate the ionization energy.

Step 2: Recall the formula for the energy of an electron in the Bohr model: $ E_n = - \frac{Z_{\text{eff}}^2 \cdot e^4}{8 \cdot \pi^2 \cdot \epsilon_0^2 \cdot h^2 \cdot n^2} $, where $ Z_{\text{eff}} $ is the effective nuclear charge, $ e $ is the elementary charge, $ \epsilon_0 $ is the permittivity of free space, $ h $ is Planck's constant, and $ n $ is the principal quantum number.

Step 3: Substitute the given values into the formula. For rubidium's 5s electron, $ Z_{\text{eff}} = 2.771 $, $ n = 5 $, and the constants $ e $, $ \epsilon_0 $, and $ h $ are known physical constants. Ensure all units are consistent (e.g., SI units).

Step 4: Simplify the expression to calculate the energy $ E_n $. This will give the energy of the electron in joules. Since ionization energy is the energy required to remove the electron, take the absolute value of $ E_n $.

Step 5: Convert the ionization energy from joules to electron volts (eV) using the conversion factor $ 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} $. This will provide the ionization energy in a more convenient unit for atomic-scale calculations.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Effective Nuclear Charge

Effective nuclear charge (Z_eff) is the net positive charge experienced by an electron in a multi-electron atom. It accounts for the shielding effect of inner electrons, which reduces the full nuclear charge felt by outer electrons. In this case, the 5s electron in rubidium experiences an effective charge of 2.771e, which influences its ionization energy.

Ionization Energy

Ionization energy is the amount of energy required to remove an electron from an atom or ion in its gaseous state. It is influenced by the effective nuclear charge and the distance of the electron from the nucleus. Higher effective nuclear charge typically results in higher ionization energy, as electrons are held more tightly by the nucleus.

Hydrogen-like Atom Model

The hydrogen-like atom model simplifies the calculation of ionization energy for multi-electron atoms by treating them as if they were similar to hydrogen, where only one electron is present. The formula for ionization energy can be adapted from the hydrogen atom's energy levels, using the effective nuclear charge to determine the energy required to remove the electron from the atom.

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