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Chapter 27: Problem 43

Magnetic Moment of the Hydrogen Atom. In the Bohr model of the hydrogen atom (see Section \(38.5 ),\) in the lowest energy state the electron orbits the proton at a speed of \(2.2 \times\) \(10^{6} \mathrm{m} / \mathrm{s}\) in a circular orbit of radius \(5.3 \times 10^{-11} \mathrm{m}\) (a) What is the orbital period of the electron? (b) If the orbiting electron is considered to be a current loop, what is the current \(I ?\) (c) What is the magnetic moment of the atom due to the motion of the electron?

Short Answer

Expert verified
(a) \(1.52 \times 10^{-16} \text{ s}\), (b) \(1.05 \times 10^{-3} \text{ A}\), (c) \(9.24 \times 10^{-24} \text{ A} \cdot \text{m}^2\).

Step by step solution

01

Identify Given Values

In the Bohr model, we have the following known values for the electron in the lowest energy state: - Speed of the electron, \( v = 2.2 \times 10^6 \text{ m/s} \)- Radius of the orbit, \( r = 5.3 \times 10^{-11} \text{ m} \)
02

Calculate Orbital Period

The orbital period \( T \) is the time it takes for the electron to complete one full orbit. It can be determined using the formula \( T = \frac{2\pi r}{v} \). Substituting the known values:\[ T = \frac{2\pi \times 5.3 \times 10^{-11} \text{ m}}{2.2 \times 10^6 \text{ m/s}} \]Calculate to find:\[ T \approx 1.52 \times 10^{-16} \text{ s} \]
03

Determine the Current

The current \( I \) in a current loop caused by the moving electron can be calculated with \( I = \frac{e}{T} \), where \( e \) is the elementary charge \( e = 1.6 \times 10^{-19} \text{ C} \). Using the period calculated in the previous step:\[ I = \frac{1.6 \times 10^{-19} \text{ C}}{1.52 \times 10^{-16} \text{ s}} \]Calculate to find:\[ I \approx 1.05 \times 10^{-3} \text{ A} \]
04

Calculate Magnetic Moment

The magnetic moment \( \mu \) for the loop is given by \( \mu = IA \), where \( A \) is the area of the loop \( A = \pi r^2 \). Calculate the area:\[ A = \pi \times (5.3 \times 10^{-11} \text{ m})^2 \approx 8.8 \times 10^{-21} \text{ m}^2 \]Now calculate the magnetic moment:\[ \mu = 1.05 \times 10^{-3} \times 8.8 \times 10^{-21} \approx 9.24 \times 10^{-24} \text{ A} \cdot \text{m}^2 \]

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Key Concepts

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

Magnetic Moment
The magnetic moment forms a crucial part of understanding magnetic fields generated by moving charges, such as electrons. In the Bohr model of the hydrogen atom, the magnetic moment (\( \mu \) ) arises because the electron orbits the proton. As the electron moves, it creates a loop current, generating a magnetic field. This field's strength and direction depend on the magnetic moment. The magnetic moment is calculated using the formula:
  • \( \mu = I \times A \)
where \( I \) is the current, and \( A \) is the area of the electron's orbit.
To find \( \mu \), you first determine the current, using the elementary charge over the orbital period. Then, find the area of the orbit by using the radius of the electron鈥檚 path, \( r \), in the equation \( A = \pi r^2 \). This calculation is essential in quantum physics for understanding atomic behaviors and interactions.
Current Loop
A current loop refers to a setup where moving charges form a closed path, generating a magnetic field. In the Bohr model, the electron revolving around the proton acts like a current loop. As the electron moves at a constant speed along its circular path, it creates an electrical current. This is because the circulating electron behaves like a tiny loop of current due to its perpetual motion in orbit.
The current (\( I \) ) is calculated using:
  • \( I = \frac{e}{T} \)
where (\( e \) ) is the elementary charge of the electron, and (\( T \) ) is the orbital period which is the time taken for one complete orbit. This current is small but significant enough to generate a magnetic field, influencing the magnetic properties of atoms.
Orbital Period
The orbital period represents the time it takes for the electron to complete one full journey around the nucleus. In the context of the Bohr model, determining this period helps in understanding how quickly the electron orbits in its path around the proton. It's a critical concept, as the frequency of orbit relates directly to the electron's energy state.
  • The orbital period, \( T \), is found using the formula: \( T = \frac{2\pi r}{v} \)
where \( r \) is the radius of the orbit, and \( v \) is the speed of the electron. Calculating the orbital period provides insights into the orbit dynamics and is key in any further calculations involving the magnetic moment and current loop.
Elementary Charge
The elementary charge is the constant that represents the amount of electric charge carried by a single proton or electron. Understanding this charge is essential in calculations involving electrical phenomena on atomic levels. The value is approximately \( 1.6 \times 10^{-19} \text{ C} \).
In the Bohr model, this value is pivotal when calculating the current (\( I \) ). This is because the moving electron, acting as a current loop, has its current deduced through:
  • \( I = \frac{e}{T} \)
where \( T \) is the previously calculated orbital period. Recognizing the role of this charge allows a deeper insight into how atomic particles interact electrically and magnetically, forming the foundational understanding of atomic structures and interactions.

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Most popular questions from this chapter

A particle with mass \(1.81 \times 10^{-3} \mathrm{kg}\) and a charge of \(1.22 \times\) \(10^{-8} \mathrm{C}\) has, at a given instant, a velocity \(\vec{v}=\left(3.00 \times 10^{4} \mathrm{m} / \mathrm{s}\right) \hat{j}\) What are the magnitude and direction of the particle's accoleration produced by a uniform magnetic field \(\vec{B}=(1.63 \mathrm{T}) \hat{\imath}+\) \((0.980 \mathrm{T}) \hat{\jmath} ?\)

(a) An 16 nucleus (charge \(+8 e )\) moving horizontally from west to east with a speed of 500 \(\mathrm{km} / \mathrm{s}\) experiences a magnetic force of 0.00320 \(\mathrm{nN}\) vertically downward. Find the magnitude and direction of the weakest magnetic field required to produce this force. Explain how this same force could be caused by a larger magnetic field. (b) An electron moves in a uniform, horizontal, 2.10 - T magnetic field that is toward the west. What must the magnitude and direction of the minimum velocity of the electron be so that the magnetic force on it will be 4.60 \(\mathrm{pN}\) , vertically upward? Explain how the velocity could be greater than this minimum value and the force still have this same magnitude and direction.

Using Gauss's Law for Magnetism. In a certain region of space, the magnetic field \(\overrightarrow{\boldsymbol{B}}\) is not uniform. The magnetic field has both a \(z\) -component and a component that points radially away from or toward the \(z\) -axis. The z-component is given by \(B_{z (z)=\beta z,\) where \(\beta\) is a positive constant. The radial component \(B_{x}\) depends only on \(r\) , the radial distance from the \(z\) -axis. (a) Use Gauss's law for magnetism, Eq. \((27.8),\) to find the radial component \(B_{r}\) as a function of \(r .\) (Hint Try a cylindrical Gaussian surface of radius \(r\) concentric with the \(z\) -axis, with one end at \(z=0\) and the other at \(z=L . )\) (b) Sketch the magnetic field lines.

A particle with charge \(-5.60 \mathrm{nC}\) is moving in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=-(1.25 \mathrm{T}) \hat{\boldsymbol{k}} .\) The magnetic force on the particle is measured to be \(\overrightarrow{\boldsymbol{F}}=-\left(3.40 \times 10^{-7} \mathrm{N}\right) \hat{\boldsymbol{i}}+\left(7.40 \times 10^{-7} \mathrm{N}\right) \hat{\boldsymbol{j}}\) (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\overrightarrow{\boldsymbol{F}} \cdot \overrightarrow{\boldsymbol{F}}\) . What is the angle between \(\overrightarrow{\boldsymbol{v}}\) and \(\overrightarrow{\boldsymbol{F}} ?\)

A \(150-\) g ball containing \(4.00 \times 10^{8}\) excess electrons is dropped into a \(125-\mathrm{m}\) vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.250 \(\mathrm{T}\) and direction from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enrers the field.
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