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The energy of the \(n=2\) Bohr orbit is \(-30.6 \mathrm{eV}\) for an unidentified ionized atom in which only one electron moves about the nucleus. What is the radius of the \(n=5\) orbit for this species?

Short Answer

Expert verified
The radius of the \( n=5 \) orbit is calculated using the Bohr formula as \( r_5 = \frac{5^2 \times 0.529}{Z} \ \mathrm{\AA} \).

Step by step solution

01

Understand the Formula for Bohr Radius

The Bohr radius equation for an orbit in an ionized atom is given by \( r_n = \frac{n^2 h^2}{4\pi^2 m k_0 Z e^2} \). Here, \( n \) is the orbit number, \( h \) is Planck's constant, \( m \) is the electron mass, \( k_0 \) is the Coulomb's constant, \( Z \) is the atomic number, and \( e \) is the elementary charge.
02

Relate Energy to Atomic Number (Z)

The energy for a level \( n \) is given by \( E_n = -\frac{Z^2 R_H}{n^2} \), where \( R_H \) is the Rydberg constant. For \( n=2 \), given \( E_2 = -30.6 \mathrm{eV} \), we can solve for \( Z \) using \( -30.6 = -\frac{Z^2 R_H}{2^2} \).
03

Calculate Z

Rearrange the energy formula to find \( Z^2 = \frac{4 \times 30.6}{R_H} \) and compute \( Z \) from this equation using \( R_H = 13.6 \mathrm{eV} \).
04

Apply Bohr's Radius Formula for n=5

With \( Z \) found, apply the Bohr radius formula \( r_n = \frac{n^2 a_0}{Z} \) for \( n=5 \), where \( a_0 = 0.529 \mathrm{\AA} \) is the Bohr radius for hydrogen.
05

Calculate the Radius

Calculate \( r_5 \) using \( r_5 = \frac{5^2 \times 0.529}{Z} \) \( \mathrm{\AA} \) using the value of \( Z \) obtained in Step 3.

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Key Concepts

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

Bohr radius
To understand the Bohr radius, think of it as the fundamental distance from the nucleus to the electron in the simplest hydrogen atom, at its lowest energy level. The Bohr radius is a constant, denoted as \( a_0 \), and is approximately 0.529 Ã…. This concept is crucial as it helps calculate the size of any orbit in an atom, particularly those in excited states, like \( n=5 \).

When dealing with ionized atoms, the Bohr radius formula must be adjusted to account for different atomic numbers \( Z \). The generalized formula becomes:
  • \( r_n = \frac{n^2 a_0}{Z} \)
Here, \( n \) represents the orbit number, and \( Z \) is the atomic number which affects the orbit size. Larger \( Z \) values mean stronger nuclear attraction, leading to smaller orbits for similar \( n \).
ionized atom
An ionized atom is one in which electrons have been removed or added, giving the atom a net positive or negative charge. In the context of the exercise, we are considering an ionized atom with a single electron. This situation closely resembles a hydrogen-like atom, allowing the use of Bohr's model to determine energy levels and radii.

When an atom loses an electron, it becomes positively charged, and the positive charge of the nucleus now exerts a stronger pull on the remaining electron. This change affects the electron's energy levels and orbit paths. The higher the charge (or the atomic number \( Z \)), the more tightly the remaining electron is bound, affecting the calculation of the electron's orbit size. Understanding this is crucial for accurately applying the Bohr model in calculations.
energy levels
Energy levels in an atom are the fixed distances from the nucleus where electrons are likely to be found. Each level corresponds to a specific amount of energy. The Bohr model simplifies this by considering electrons in defined circular orbits. The energy of any level \( n \) is given by the formula
  • \( E_n = -\frac{Z^2 R_H}{n^2} \)
Where \( R_H \) is the Rydberg constant, and \( Z \) is the atomic number.

In our exercise, knowing the energy of the \( n=2 \) orbit helps us determine \( Z \), showing us how closely tied energy levels are to other properties. Higher energy levels (larger \( n \)) mean that the electron occupies a higher orbit, which is less tightly bound and generally larger, if the atom is not ionized further.
atomic number
The atomic number \( Z \) of an element denotes the number of protons in its nucleus. In the Bohr model, \( Z \) determines the attractive force between the nucleus and electrons, influencing the electron's orbits and energies.

This number is particularly pertinent when working with ionized atoms, as it dictates how the electron behaves differently compared to in a hydrogen atom. For example, higher \( Z \) values mean more protons, which increases the electromagnetic force exerted on the electron.

In the exercise, determining \( Z \) is necessary to apply formulas correctly. By solving for \( Z \), you calibrate the model to predict the energies and radii of orbits for the atom in question. Thus, \( Z \) serves as the linchpin in linking atomic structure to observed physical properties of the atom.

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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 \(\ell\) 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) \(2 p \rightarrow 1 s\) (c) \(4 p \rightarrow 2 p(d) 4 s \rightarrow 2 p\) (e) \(3 \mathrm{d} \rightarrow 3 \mathrm{s}\)

A hydrogen atom is in the ground state. It absorbs energy and makes a transition to the \(n=3\) excited state. The atom returns to the ground state by emitting two photons. What are their wavelengths?

The electron in a certain hydrogen atom has an angular momentum of \(8.948 \times 10^{-34} \mathrm{J} \cdot \mathrm{s} .\) What is the largest possible magnitude for the \(z\) component of the angular momentum of this electron? For accuracy, use \(h=6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}\).

Multiple-Concept Example 9 reviews the concepts that are important in this problem. An electron, traveling at a speed of \(6.00 \times 10^{7} \mathrm{m} / \mathrm{s},\) strikes the target of an X-ray tube. Upon impact, the electron decelerates to one-quarter of its original speed, an X-ray photon being emitted in the process. What is the wavelength of the photon?

A hydrogen atom emits a photon that has momentum with a magnitude of \(5.452 \times 10^{-27} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\). This photon is emitted because the electron in the atom falls from a higher energy level into the \(n=1\) level. What is the quantum number of the level from which the electron falls? Use a value of \(6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}\) for Planck's constant.

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