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Chapter 27: Problem 50

Magnetic Moment of the Hydrogen Atom. In the Bohr model of the hydrogen atom (see Section 39.3 ), 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
The orbital period of the electron is calculated as \(T\), the current produced due to the orbiting electron is calculated as \(I\) and finally, using \(I\), the magnetic moment of the atom due to the motion of the electron is calculated as \(\mu\).

Step by step solution

01

Determine the Orbital Period

To find the orbital period \(T\), we can use the relationship between speed, distance and time given by the formula: \(speed = \frac{distance}{time}\). Rearranging the equation for \(time\), we get \(T = \frac{distance}{speed}\). Here, the distance is the circumference of the circular path and can be calculated as \(2\pi r\) where \(r = 5.3 \times 10^{-11}\ m\). Substituting \(r\) and the given speed \(v = 2.2 \times 10^{6}\ m/s\) we can calculate \(T\).
02

Calculate the Current

The current \(I\) is produced due to the orbiting electron which completes one revolution in a time \(T\). The amount of charge \(Q\) transferred in one complete orbit is equal to the magnitude of the electronic charge \(Q = 1.6 \times 10^{-19} C\). Using the relationship \(I = \frac{Q}{T}\), we can substitute \(Q\) and \(T\) to calculate \(I\).
03

Determine the Magnetic Moment

The magnetic moment \(\mu\) due to the motion of the electron is given by the relationship \(\mu = I \times A\) where \(A\) is the area of the circular path. We can calculate \(A\) using \(\pi r^{2}\) and substituting the value of \(I\) from the previous step along with \(A\), we can get \(\mu\).

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Key Concepts

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

Bohr Model
The Bohr model of the atom was a pivotal stepping stone in the development of atomic theory. Proposed by Niels Bohr in 1913, this model describes the atom as a small, positively charged nucleus surrounded by electrons that travel in circular orbits around the nucleus鈥攎uch like planets orbiting the Sun, but with attraction provided by electrostatic forces rather than gravity.

The Key Postulates

  • Electrons move in prescribed orbits or 'shells' with fixed energies.
  • Energy is only absorbed or emitted when an electron moves between these orbits.
These postulates, while having limitations, provided a good explanation for the hydrogen atom's spectrum and laid the groundwork for the more complex models that followed.
Orbital Period
The orbital period in the context of an electron within an atom is the time it takes for the electron to make one complete revolution around the nucleus. Calculating this involves concepts from classical mechanics, despite the quantum mechanical nature of electrons in atoms. The speed of the electron and the circumference of its orbit (determined by its radius) factor into finding the orbital period.

Understanding Orbital Period in Bohr's Model

  • The circumference of the electron's orbit is given by the formula: \( 2\pi r \).
  • Given the speed, we can calculate the period (\( T \)) with \( T = \frac{2\pi r}{v} \), where \( r \) is the radius and \( v \) is the speed of the electron.
Insights into the orbital period of an electron allow us to better understand electron transitions and the absorption or emission of light.
Circular Motion
Circular motion for an electron orbiting a nucleus involves moving along a circular path. This type of motion is characterized by the electron maintaining a constant speed while changing direction at every point along its path鈥攔esulting in a constant change in velocity (since velocity is a vector quantity).

Features of Circular Motion

  • Constant speed with a continuously changing direction.
  • The net force (centripetal force) is directed towards the center of the circular path.
  • The centripetal force in the Bohr model is the electrostatic force of attraction between the electron and the nucleus.
Circular motion is a fundamental concept not only in orbiting electrons but in many areas of physics, including planetary motion and machinery.
Magnetic Moment
A magnetic moment is a vector quantity that represents the magnetic strength and orientation of a magnetic object or particle. In an atom, it's caused by the motion of electrons, both due to their orbital movements and their intrinsic spin. For the hydrogen atom in its lowest energy state, the magnetic moment can be found due to its orbiting electron acting like a tiny current loop.

Determining Magnetic Moment

  • It's calculated by multiplying the current \( I \) with the area \( A \) of the electron's orbit: \( \mu = I \times A \).
  • The area is found using the formula for the area of a circle: \( A = \pi r^{2} \), where \( r \) is the radius.
  • The current \( I \) is the charge of the electron divided by the orbital period, or \( I = \frac{Q}{T} \).
The magnetic moment is a crucial aspect of understanding how atoms interact with magnetic fields, which has important implications in fields such as magnetic resonance imaging (MRI) and other technologies.

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Most popular questions from this chapter

A Cycloidal Path. A particle with mass \(m\) and positive charge \(q\) starts from rest at the origin shown in Fig. \(\mathbf{P 2 7 . 8 2 .}\). There is a uniform electric field \(\vec{E}\) in the \(+y\) -direction and a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}\) directed out of the page. It is shown in more advanced books that the path is a cycloid whose radius of curvature at the top points is twice the \(y\) -coordinate at that level. (a) Explain why the path has this general shape and why it is repetitive. (b) Prove that the speed at any point is equal to \(\sqrt{2 q E y / m}\). (Hint: Use energy conservation.) (c) Applying Newton's second law at the top point and taking as given that the radius of curvature here equals \(2 y,\) prove that the speed at this point is \(2 E / B\)

In a 1.25 T magnetic field directed vertically upward, a particle having a charge of magnitude \(8.50 \mu \mathrm{C}\) and initially moving northward at \(4.75 \mathrm{~km} / \mathrm{s}\) is deflected toward the east. (a) What is the sign of the charge of this particle? Make a sketch to illustrate how you found your answer. (b) Find the magnetic force on the particle.

A particle of mass \(0.195 \mathrm{~g}\) carries a charge of \(-2.50 \times 10^{-8} \mathrm{C}\). The particle is given an initial horizontal velocity that is due north and has magnitude \(4.00 \times 10^{4} \mathrm{~m} / \mathrm{s}\). What are the magnitude and direction of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction?

A straight, vertical wire carries a current of 2.60 A down- ward in a region between the poles of a large superconducting electromagnet, where the magnetic field has magnitude \(B=0.588 \mathrm{~T}\) and is horizontal. What are the magnitude and direction of the magnetic force on a \(1.00 \mathrm{~cm}\) section of the wire that is in this uniform magnetic field, if the magnetic field direction is (a) east; (b) south; (c) \(30.0^{\circ}\) south of west?

An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of \(6.64 \times 10^{-27} \mathrm{~kg}\) ) traveling horizontally at \(35.6 \mathrm{~km} / \mathrm{s}\) enters a uniform, vertical, \(1.80 \mathrm{~T}\) magnetic field. (a) What is the diameter of the path followed by this alpha particle? (b) What effect does the magnetic field have on the speed of the particle? (c) What are the magnitude and direction of the acceleration of the alpha particle while it is in the magnetic field? (d) Explain why the speed of the particle does not change even though an unbalanced external force acts on it.
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