Problem 2 Consider an electron in the firs... [FREE SOLUTION]

Chapter 3: Problem 2

Consider an electron in the first Bohr orbit of a hydrogen atom. (a) What is the radius (in meters) of this orbit? (b) What is the total energy (in eV) of the electron in this orbit? (c) How much energy is required to ionize a hydrogen atom when the electron is in the ground state?

Short Answer

Expert verified

(a) Radius = \( 5.29 \times 10^{-11} \) m, (b) Total energy = -13.6 eV, (c) Ionization energy = 13.6 eV.

Step by step solution

- Understanding the Planck Constant and Bohr Radius

To determine the radius of the first Bohr orbit, use the Bohr radius formula: \[ r_n = \frac{n^2 \times h^2 \times \text{蔚}_0}{\text{蟺} \times m_e \times e^2} \]Where: \(n\) is the orbit number (1 for the first orbit), \(h\) is Planck's constant (\(6.626 \times 10^{-34} \text{J} \text{s}\)), \( \text{蔚}_0 \) is the permittivity of free space (\(8.85 \times 10^{-12} \frac{C^2}{N\text{m}^2}\)), \(m_e\) is the mass of an electron (\(9.109 \times 10^{-31} \text{kg}\)), \(e\) is the elementary charge (\(1.602 \times 10^{-19} \text{C}\)).

- Calculate the Radius of the First Bohr Orbit

Substitute the known values into the Bohr radius formula for the first orbit (\(n = 1\)): \[ r_1 = \frac{1^2 \times (6.626 \times 10^{-34})^2 \times (8.85 \times 10^{-12})}{\text{蟺} \times (9.109 \times 10^{-31}) \times (1.602 \times 10^{-19})^2} \]Solving this yields \[ r_1 鈮� 5.29 \times 10^{-11} \text{m} \]

- Determine the Total Energy of the Electron in the First Orbit

Use the formula for the energy of an electron in the nth Bohr orbit: \[ E_n = - \frac{13.6 \text{eV}}{n^2} \]For the first orbit (\(n = 1\)): \[ E_1 = - \frac{13.6 \text{eV}}{1^2} = -13.6 \text{eV} \]

- Calculate the Ionization Energy of the Hydrogen Atom

Ionization energy is the energy required to remove the electron completely from the atom. For an electron in the ground state (first orbit), this is the same as the total energy with opposite sign: \[ E_{\text{ionization}} = 13.6 \text{eV} \]

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Key Concepts

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

Bohr radius

The Bohr radius is a fundamental concept in atomic physics. It represents the average distance between the electron and the nucleus in the ground state of a hydrogen atom. For any given orbit, the radius can be calculated using the Bohr radius formula:
\[ r_n = \frac{n^2 \times h^2 \times \epsilon_0}{\pi \times m_e \times e^2} \]
Here, each variable has a specific meaning:

\(n\) is the orbit number.
\(h\), Planck's constant, is \(6.626 \times 10^{-34} \text{Js}\).
\(\epsilon_0\), the permittivity of free space, is \(8.85 \times 10^{-12} \frac{C^2}{Nm^2}\).
\(m_e\), the electron mass, is \(9.109 \times 10^{-31} \text{kg}\).
\(e\), the elementary charge, is \(1.602 \times 10^{-19} \text{C}\).

When you substitute these values to find the radius of the first orbit ( = 1), you get \[ r_1 鈮� 5.29 \times 10^{-11} \text{m} \]
This is a fundamental constant known as the Bohr radius.

Electron energy levels

In the Bohr model of the hydrogen atom, electrons occupy specific energy levels or orbits around the nucleus. These orbits are quantized, meaning electrons can only exist in certain well-defined energy levels. The energy of an electron in any orbit is given by the formula:
\[ E_n = - \frac{13.6 \text{eV}}{n^2} \]
In this equation:

\(E_n\) is the energy of the electron in the nth orbit.
13.6 eV is the energy of the electron in the ground state (first orbit).
\(n\) is the orbit number.

For the first orbit ( = 1), the energy is \( E_1 = -13.6 \text{ eV} \). This negative sign indicates that the electron is bound to the nucleus and would need 13.6 eV of energy to be freed.
As \(n\) increases, the energy becomes less negative, meaning the electron is less tightly bound. For higher orbits, the electron is farther from the nucleus, making it easier to remove.

Ionization energy

Ionization energy is the amount of energy required to remove an electron completely from an atom. For a hydrogen atom, when the electron is in the ground state, this becomes crucial in understanding atomic behavior. The ionization energy formula is straightforward:

\( E_{\text{ionization}} = 13.6 \text{ eV} \)

This means to ionize a hydrogen atom (remove its electron completely), you need to provide 13.6 eV of energy.

When the electron is in the first orbit (ground state), its energy is \(-13.6 \text{eV}\). To ionize the atom, you need to overcome this binding energy, hence the ionization energy is \( +13.6 \text{eV}\).
This value changes if the electron is in a higher orbit. For example, if in the second orbit ( = 2), the energy would be \( E_2 = - \frac{13.6}{4} = -3.4 \text{eV}\), so it would take 3.4 eV to ionize it from the second orbit.

Understanding these concepts can help unravel many principles in atomic physics and chemistry.

91影视

Consider an electron in the first Bohr orbit of a hydrogen atom. (a) What is the radius (in meters) of this orbit? (b) What is the total energy (in eV) of the electron in this orbit? (c) How much energy is required to ionize a hydrogen atom when the electron is in the ground state?

Short Answer

Step by step solution

- Understanding the Planck Constant and Bohr Radius

- Calculate the Radius of the First Bohr Orbit

- Determine the Total Energy of the Electron in the First Orbit

- Calculate the Ionization Energy of the Hydrogen Atom

Key Concepts

Bohr radius

Electron energy levels

Ionization energy

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