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A singly ionized helium atom (He') has only one electron in orbit about the nucleus. What is the radius of the ion when it is in the second excited state?

Short Answer

Expert verified
The radius in the second excited state is approximately 2.38 x 10^{-10} meters.

Step by step solution

01

Identify Key Information

The problem involves a helium ion, specifically a singly ionized helium atom ( He^{+} ), which has one electron orbiting the nucleus, similar to hydrogen but with two protons in the nucleus. The question asks for the radius of the ion in the second excited state, which corresponds to the principal quantum number n = 3 .
02

Apply Bohr's Model for Radius

In Bohr's model for hydrogen-like atoms, the radius of an orbit for any energy level n is given by: R_n = n^2 imes a_0 / Z , where a_0 is the Bohr radius ( 0.529 imes 10^{-10} m), n is the principal quantum number, and Z is the atomic number. For helium ion, Z = 2 .
03

Calculate the Radius

Substitute n = 3 , Z = 2 , and a_0 = 0.529 imes 10^{-10} m into the formula: R_n = (3^2) imes 0.529 imes 10^{-10} ext{ m} / 2 = 2.38 imes 10^{-10} ext{ m}.
04

Conclude with the Solution

The radius of the singly ionized helium atom ( He^{+} ) in the second excited state ( n = 3 ) is approximately 2.38 imes 10^{-10} meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Singly Ionized Helium Atom
A singly ionized helium atom, denoted as He鈦, is a helium atom that has lost one of its two electrons. This results in a positively charged ion with just a single electron orbiting around the nucleus. In essence, this configuration mirrors a hydrogen atom, where there is one electron and one proton, but in the helium ion case, two protons are present in the nucleus. This distinction is important because the presence of two protons impacts the atom's behavior according to Bohr's model. This increase in the nuclear charge compared to hydrogen significantly affects the electron's orbit and energy levels, leading to a smaller orbit at similar energy levels compared to hydrogen. You'll often encounter singly ionized helium atoms in contexts similar to hydrogen due to their hydrogen-like simplicity but with intriguing differences.
Principal Quantum Number
The principal quantum number, denoted by the symbol \( n \), plays a crucial role in the Bohr model of atomic structure. It dictates the energy level of an electron within an atom. In essence, it defines the electron's shell. The higher the value of \( n \), the greater the distance of the electron's orbit from the nucleus and the higher its energy. For example, when an electron is excited to higher energy levels, it moves to orbits with larger \( n \). This is why the second excited state of an electron refers to \( n = 3 \). It's important because a change in the principal quantum number affects the radius of the electron's orbit and its energy. In hydrogen-like systems such as the singly ionized helium atom, \( n \) determines how the atom behaves in different states.
Atomic Number
The atomic number, symbolized as \( Z \), tells us the number of protons in the nucleus of an atom. This fundamental concept determines the element's identity and its position in the periodic table. For helium, the atomic number is 2, meaning every helium atom has 2 protons. In relation to the Bohr model, the atomic number is crucial because it determines the strength of the nuclear charge, which in turn affects the electron's orbit in hydrogen-like atoms. For the singly ionized helium atom, \( Z = 2 \), so when using Bohr's equation, this factor reduces the radius of the electron's orbit as compared to hydrogen, which has \( Z = 1 \). Therefore, the atomic number directly influences how tightly the electron is bound to the nucleus and the dimensions of its path.
Bohr Radius
The Bohr radius, represented by \( a_0 \), is a fundamental physical constant that represents the most probable distance between the nucleus and the orbiting electron in a hydrogen atom in its ground state. It measures approximately \( 0.529 \times 10^{-10} \) meters. This concept is particularly useful in Bohr's model for describing hydrogen-like atoms. Since the Bohr radius sets the scale for the size of electron orbits, it serves as a baseline for calculating radii of orbits at higher energy levels in different elements. When computing the radius of an electron's orbit in a hydrogen-like atom, including a singly ionized helium atom, the Bohr radius is pivotal. The formula \( R_n = n^2 \frac{a_0}{Z} \) uses \( a_0 \) together with the principal quantum number and the atomic number to determine the orbit size, illustrating the Bohr radius's central role in the model.

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Most popular questions from this chapter

When an electron makes a transition between energy levels of an atom, there are no restrictions on the initial and final values of the principal quantum number \(n .\) According to quantum mechanics, however, there is a rule that restricts the initial and final values of the orbital quantum number \(\ell\). This rule is called a selection rule and states that \(\Delta \ell=\pm 1 .\) In other words, when an electron makes a transition between energy levels, the value of \(t\) can only increase or decrease by one. The value of \(\ell\) may not remain the same nor may it increase or decrease by more than one. According to this rule, which of the following energy level transitions are allowed? (a) \(2 \mathrm{s} \rightarrow 1 \mathrm{s}\) (b) \(2 p \rightarrow\) Is \((c) 4 p \rightarrow 2 p\) (d) \(4 \mathrm{s} \rightarrow 2 \mathrm{p}\) (e) \(3 \mathrm{d} \rightarrow 3 \mathrm{s}\)

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(a) What is the ionization energy of a hydrogen atom that is in the \(n=4\) excited state? (b) For a hydrogen atom, determine the ratio of the ionization energy for the \(n=4\) excited state to the ionization energy for the ground state.

There are \(Z\) protons in the nucleus of an atom, where \(Z\) is the atomic number of the element. An \(\alpha\) particle carries a charge of \(+2 \mathrm{e}\) In a scattering experiment, an \(\alpha\) particle, heading directly toward a nucleus in a metal foil, will come to a halt when all the particle's kinetic energy is converted to electric potential energy. In such a situation, how close will an \(\alpha\) particle with a kinetic energy of \(5.0 \times 10^{-13} \mathrm{J}\) come to a gold nucleus \((Z=79) ?\)

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