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Chapter 27: Problem 54

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\(^6\) m/s in a circular orbit of radius 5.3 \(\times\) 10\(^{-11}\) 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 脳 10鈦宦光伓 s; (b) 1.05 脳 10鈦宦� A; (c) 9.27 脳 10鈦宦测伌 J/T.

Step by step solution

01

Understanding Orbital Period

The orbital period of the electron is the time it takes for the electron to complete one full orbit around the proton. This is given by the formula \( T = \frac{2\pi r}{v} \), where \( r \) is the radius of the orbit and \( v \) is the speed of the electron. We have \( r = 5.3 \times 10^{-11} \text{ m} \) and \( v = 2.2 \times 10^6 \text{ m/s} \).
02

Calculate Orbital Period

Substitute the values into the formula:\[T = \frac{2\pi \times 5.3 \times 10^{-11} \text{ m}}{2.2 \times 10^6 \text{ m/s}}.\] Calculate to find:\[T \approx 1.52 \times 10^{-16} \text{ seconds}.\]
03

Understanding Current in the Orbit

If the orbiting electron is considered as a loop of current, its current \( I \) is given by \( I = \frac{e}{T} \), where \( e \) is the elementary charge \( 1.6 \times 10^{-19} \text{ C} \) and \( T \) is the period of the electron's orbit. Use the value of \( T \) from the previous step.
04

Calculate the Current

Substitute the values into the current formula:\[I = \frac{1.6 \times 10^{-19} \text{ C}}{1.52 \times 10^{-16} \text{ s}}\]Calculate to find:\[I \approx 1.05 \times 10^{-3} \text{ A}.\]
05

Understanding Magnetic Moment

The magnetic moment \( \mu \) due to the orbit of the electron is given by the formula \( \mu = I \cdot A \), where \( A = \pi r^2 \) is the area of the orbit. Use the previous result for \( I \) and \( r = 5.3 \times 10^{-11} \text{ m} \).
06

Calculate Magnetic Moment

Substitute the values into the magnetic moment formula:\[A = \pi \cdot (5.3 \times 10^{-11})^2\]\[\mu = 1.05 \times 10^{-3} \cdot \pi \cdot (5.3 \times 10^{-11})^2\]Calculate to find:\[\mu \approx 9.27 \times 10^{-24} \text{ J/T}.\]

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Key Concepts

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

Orbital Period of an Electron
The concept of the orbital period in the context of the Bohr model revolves around understanding the time it takes for an electron to complete one full orbit around the nucleus. This analogy is similar to the time Earth takes to revolve around the Sun in its orbit. In the Bohr model, the electron orbits the nucleus at a constant speed, making calculations straightforward. This duration is expressed by the formula:
  • \( T = \frac{2\pi r}{v} \)
where:
  • \( r \) is the radius of the orbit
  • \( v \) is the velocity of the electron
For example, with a radius of \( r = 5.3 \times 10^{-11} \) meters and a speed of \( v = 2.2 \times 10^6 \) meters per second, the calculation becomes simple.
The result is approximately \( 1.52 \times 10^{-16} \) seconds, showcasing the incredibly brief time an electron takes to complete its journey around the nucleus.
Current in Electron Orbit
In the Bohr model, the orbiting electron can be thought of as creating a tiny loop of current due to its constant motion around the nucleus. Although it might sound abstract, this approximation is essential for understanding magnetic properties in atomic structures. The current \( I \) generated by this motion is quantified by:
  • \( I = \frac{e}{T} \)
where:
  • \( e \) is the elementary charge, valued at \( 1.6 \times 10^{-19} \) Coulombs
  • \( T \) is the orbital period just calculated
Using the orbital period from earlier, the current is calculated as approximately \( 1.05 \times 10^{-3} \) Amperes.
By conceptualizing the electron's motion as a current loop, we get a foundational understanding of how electrons contribute to magnetic fields in atomic-scale systems.
Magnetic Moment of an Atom
The magnetic moment within the Bohr model is a key concept in grasping how atoms respond to magnetic fields. This moment arises due to the orbiting electron's motion, similar to how a current loop generates a magnetic field. The magnetic moment \( \mu \) is calculated by the formula:
  • \( \mu = I \cdot A \)
where:
  • \( I \) is the current from the electron's orbit
  • \( A \) is the area of the orbit, \( A = \pi r^2 \)
Utilizing the radius \( r = 5.3 \times 10^{-11} \) meters, this calculation yields a magnetic moment of roughly \( 9.27 \times 10^{-24} \) Joules per Tesla.
This value aids in explaining how each electron within an atom contributes to magnetic phenomena, crucial for understanding magnetism at the atomic level.

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Most popular questions from this chapter

An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of 6.64 \(\times\) 10\(^{-27}\) kg) traveling horizontally at 35.6 km>s enters a uniform, vertical, 1.80-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.

A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34 \(\times\) 10\(^{-27}\) kg and a charge of \(+e\). The deuteron travels in a circular path with a radius of 6.96 mm in a magnetic field with magnitude 2.50 T. (a) Find the speed of the deuteron. (b) Find the time required for it to make half a revolution. (c) Through what potential difference would the deuteron have to be accelerated to acquire this speed?

A particle with charge -5.60 nC is moving in a uniform magnetic field \(\overrightarrow{B} =\) -(1.25 T)\(\hat{k}\). The magnetic force on the particle is measured to be \(\overrightarrow{F} =\) -(3.40 \(\times\) 10\(^{-7}\)N)\(\hat{\imath}\) + (7.40 \(\times\) 10\(^{-7}\)N)\(\hat{\jmath}\). (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 \(\vec{v}\) \(\cdot\) \(\overrightarrow{F}\). What is the angle between \(\vec{v}\) and \(\overrightarrow{F}\)?

The plane of a 5.0 cm \(\times\) 8.0 cm rectangular loop of wire is parallel to a 0.19-T magnetic field. The loop carries a current of 6.2 A. (a) What torque acts on the loop? (b) What is the magnetic moment of the loop? (c) What is the maximum torque that can be obtained with the same total length of wire carrying the same current in this magnetic field?

A dc motor with its rotor and field coils connected in series has an internal resistance of 3.2 \(\Omega\). When the motor is running at full load on a 120-V line, the emf in the rotor is 105 V. (a) What is the current drawn by the motor from the line? (b) What is the power delivered to the motor? (c) What is the mechanical power developed by the motor?
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