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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 of He+ in the second excited state is approximately 0.238 nm.

Step by step solution

01

Understanding the Atom

A singly ionized helium atom (He+) is similar to a hydrogen atom because it has only one electron. However, the nuclear charge is different, with a helium nucleus having two protons.
02

Identify Bohr's Model Equation

According to Bohr's model, the radius of an electron orbit in a hydrogen-like atom is given by: \[ r_n = rac{n^2 h^2}{4 \pi^2 m e^2} \frac{Z}{Z^2} = n^2 \frac{h^2}{4 \pi^2 m e Z e^2} \]Where: - \( n \) is the principal quantum number,- \( h \) is Planck's constant,- \( m \) is the electron mass,- \( e \) is the elementary charge,- \( Z \) is the atomic number of the nucleus.
03

Determine Effective Bohr Radius

For helium ion (He+), the Bohr radius formula simplifies to \[ r_n = n^2 \frac{h^2}{4 \pi^2 m e^2 Z} \]Since helium has two protons, \( Z = 2 \). Substitute \( n = 3 \) because it is the second excited state (ground state is \( n=1 \), first excited is \( n=2 \), second excited is \( n=3 \)).
04

Insert Constants and Solve

Using known constants:- Planck's constant \( h = 6.63 \times 10^{-34} \, m^2 kg/s \)- Electron mass \( m = 9.11 \times 10^{-31} \, kg \)- Elementary charge \( e = 1.60 \times 10^{-19} \, C \)- Vacuum permittivity constant \( \varepsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \, m^2) \)The Bohr radius for hydrogen \( a_0 = 0.529 \, \text{脜} \).The radius \( r_n = 3^2 \times 0.529 \, \text{脜} / 2 = \frac{9 \times 0.529}{2} \, \text{脜} = 2.3805 \, \text{脜} \).
05

Calculate the Final Radius

The radius of the singly ionized helium atom in the second excited state is approximately \[ r_3 = 2.3805 \times 10^{-10} \, \text{m} \] or 0.23805 nm.

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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 an intriguing system that provides insight into atomic behavior. While helium normally has two electrons, a singly ionized helium atom has lost one, resulting in just one electron orbiting its nucleus. This makes it similar to a hydrogen atom, given both systems have a single electron. However, the distinction lies in the nuclear charge. The helium nucleus contains two protons, whereas hydrogen has only one.

Because of this larger nuclear charge, a singly ionized helium atom experiences stronger electromagnetic attraction. This increased attraction impacts the behavior of the electron, allowing it to exhibit properties different from those found in a hydrogen atom. For instance, calculations of atomic properties such as the energy levels or orbital radius will be affected by the different atomic number or nuclear charge (Z = 2 for helium as opposed to Z = 1 for hydrogen).

To summarize, while sharing similarities with hydrogen, the presence of two protons in the nucleus of He鈦 gives rise to its distinct characteristics and renders it a hydrogen-like but uniquely interesting atomic system.
Principal Quantum Number
The principal quantum number, symbolized as 'n,' is a fundamental concept in atomic physics, playing a crucial role in defining the state of an electron in an atom. It represents the main energy level (or shell) of an electron and is always a positive integer. The principal quantum number determines the size and energy of an electron's orbit in a Bohr model, with each orbit corresponding to a different value of 'n'.

For atoms modeled similarly to the hydrogen atom, like the singly ionized helium atom, electrons occupy different energy levels. Each increase in 'n' usually signifies a higher energy level and a larger orbit. In the context of helium, when the problem specifies the second excited state, it refers to the electron being in the third energy level, where n = 3. This progression goes as follows:
  • n = 1: Ground state
  • n = 2: First excited state
  • n = 3: Second excited state
The dynamics of quantum numbers help explain why different electron transitions happen, leading to emission or absorption of spectra. Knowing the principal quantum number is a key step in calculating other atomic properties such as orbital radius or energy.
Helium Atomic Structure
When discussing helium's atomic structure, it is essential to acknowledge its position as the second element in the periodic table, with an atomic number (Z) of 2. Helium's structure typically includes two protons in the nucleus and two electrons orbiting in a stable configuration, making it a noble gas. To further explore its behavior, especially in the context of a singly ionized helium atom, imagine stripping away one electron, which offers a clearer perspective on the interactions present.

In its natural state, these two electrons occupy the first shell, characterized by n = 1, adhering to the lowest energy configuration, known as the ground state. Helium is inert in its most common form because its electrons fully occupy its only available shell under standard conditions. The simplicity of this configuration underscores helium's reputation as a chemically stable element.

Reflecting on the fundamental role of subatomic particles in the atom鈥檚 architecture helps understand how changes at this level influence behavior. In this case, with one less electron due to ionization, the remaining electron in He鈦 attracts closer to the higher positive charge of two protons, resulting in distinct atomic characteristics compared to its neutral form.

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Most popular questions from this chapter

The dye laser used in the treatment of the port-wine stain in Figure \(30-30\) (see Section 30.9) has a wavelength of \(585 \mathrm{~nm}\). A carbon dioxide laser produces a wavelength of \(1.06 \times 10^{-5} \mathrm{~m}\). What is the minimum number of photons that the carbon dioxide laser must produce to deliver at least as much or more energy to a target as does a single photon from the dye laser?

A hydrogen atom is in an excited state. (a) What is meant by the phrase "ionization energy of the excited state"? (b) Is the ionization energy of an excited state greater than, less than, or the same as the ionization energy for the ground state? Why? (a) What is the ionization energy of a hydrogen atom that is in the \(n=4\) state? (b) Determine the ratio of the ionization energy for \(n=4\) to that of the ground state. Check to make sure that your answer is consistent with your answers to the Concept Questions.

Doubly ionized lithium \(\mathrm{Li}^{2+}(Z=3)\) and triply ionized beryllium \(\mathrm{Be}^{3+}(Z=4)\) each emit a line spectrum. For a certain series of lines in the lithium spectrum, the shortest wavelength is \(40.5 \mathrm{~nm}\). For the same series of lines in the beryllium spectrum, what is the shortest wavelength?

In a Rutherford scattering experiment a target nucleus has a diameter of \(1.4 \times 10^{-14} \mathrm{~m}\). The incoming \(\alpha\) particle has a mass of \(6.64 \times 10^{-27} \mathrm{~kg} .\) What is the kinetic energy of an \(\alpha\) particle that has a de Broglie wavelength equal to the diameter of the target nucleus? Ignore relativistic effects.

A hydrogen atom is in its second excited state. Determine, according to quantum mechanics, (a) the total energy (in eV) of the atom, (b) the magnitude of the maximum angular momentum the electron can have in this state, and (c) the maximum value that the \(z\) component \(L_{7}\) of the angular momentum can have.

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