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Chapter 6: Problem 150

(a) Calculate the ionization energy for \(\mathrm{He}^{+}\). (b) What is the minimum frequency of a photon capable of ionizing \(\mathrm{He}^{+}\) ?

Short Answer

Expert verified
The ionization energy for \(\mathrm{He}^{+}\) is \(8.72 \times 10^{-18}\, \text{J}\), and the minimum frequency of a photon capable of ionizing \(\mathrm{He}^{+}\) is \(1.32 \times 10^{16}\, \text{Hz}\).

Step by step solution

01

Calculate the energy of the electron in the ground state for He+

The energy of the electron in the ground state for He+ is given by the formula: \(E_e = -\frac{13.6 \,\text{eV}}{n^2} \cdot Z^2\) Where: - \(n\) is the quantum number of the electron (1 for ground state) - \(Z\) is the atomic number of the ion (2 for He+ since it has 2 protons) Substituting the values, we get: \(E_e = -\frac{13.6\, \text{eV}}{1^2} \cdot 2^2 = -54.4\, \text{eV}\) But we need the energy in joules, so we will use the conversion factor: \(1\, \text{eV} = 1.602 \times 10^{-19}\, \text{J}\) So, \(E_e = -54.4\, \text{eV}\cdot 1.602 \times 10^{-19}\, \text{J}\, \text{eV}^{-1} = -8.72 \times 10^{-18}\, \text{J}\)
02

Calculate the ionization energy for He+

Using the ionization energy formula, we have: \(E_{ion} = -\frac{E_e}{n^2} = -\frac{-8.72 \times 10^{-18}\, \text{J}}{1^2} = 8.72 \times 10^{-18}\, \text{J}\) So, the ionization energy for He+ is \(8.72 \times 10^{-18}\, \text{J}\).
03

Calculate the minimum frequency of a photon capable of ionizing He+

Using the minimum frequency formula, we have: \(v_{min} = \frac{E_{ion}}{h} = \frac{8.72 \times 10^{-18}\, \text{J}}{6.626 \times 10^{-34}\,J\cdot s} = 1.32 \times 10^{16}\, \text{Hz}\) So, the minimum frequency of a photon capable of ionizing He+ is \(1.32 \times 10^{16}\, \text{Hz}\). In summary, the ionization energy for He+ is \(8.72 \times 10^{-18}\, \text{J}\) and the minimum frequency of a photon capable of ionizing He+ is \(1.32 \times 10^{16}\, \text{Hz}\).

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