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Chapter 19: Problem 88

In a simple model, the electron in a hydrogen atom orbits the proton at a radius of \(53 \mathrm{pm}\) and at a constant speed of $2.2 \times 10^{6} \mathrm{m} / \mathrm{s} .$ The orbital motion of the electron gives it an orbital magnetic dipole moment. (a) What is the current \(I\) in this current loop? [Hint: How long does it take the electron to make one revolution?] (b) What is the orbital dipole moment IA? (c) Compare the orbital dipole moment with the intrinsic magnetic dipole moment of the electron $\left(9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2}\right)$

Short Answer

Expert verified
Solution: (a) First, find the current I in the current loop by calculating the time taken for the electron to make one revolution and dividing the elementary charge by time. (b) Next, find the orbital dipole moment IA by multiplying the current obtained in part (a) with the area of the current loop (c) Finally, compare the values of the orbital dipole moment, IA, with the given intrinsic magnetic dipole moment of the electron, which is \(9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2}\).

Step by step solution

01

(a) Finding the current I in the current loop

First, we need to find the time taken for the electron to make one full orbit. To do that, we use the formula for the circumference of a circle and divide it by the electron's speed: Circumference = \(2\pi r\) where r is the radius of the orbit. Time taken for one revolution, T = Circumference / Speed T = \(\frac{2\pi(53 \times 10^{-12} m)}{2.2 \times 10^6 m/s}\) Now, we'll find the current using the formula: Current, I = \(\frac{Charge}{Time}\) where Charge is the elementary charge of the electron (e = \(1.6 \times 10^{-19} C\)). I = \(\frac{1.6 \times 10^{-19}C}{T}\)
02

(b) Finding the orbital dipole moment IA

The orbital dipole moment, IA, can be calculated using the current obtained in part (a) and the area of the current loop (the circle formed by the electron's movement). The formula for the area, A, of a circle is: A = \(\pi r^{2}\) So, the orbital dipole moment is: IA = \(I \times \pi (53 \times 10^{-12})^{2} m^{2}\)
03

(c) Comparing the orbital dipole moment with the intrinsic magnetic dipole moment of the electron

Now, we simply compare the value of the orbital dipole moment, IA, obtained in part (b) with the given intrinsic magnetic dipole moment of the electron, which is \(9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2}\). This will let us see how the orbital magnetic dipole moment is related to the intrinsic magnetic dipole moment of the electron.

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