Question A solid at absolute temperature \( T \) is placed in an external magnetic field \( \mathrm{H}=30,000 \) gauss. The solid contains weakly interacting paramagnetic atoms of spin-1/2 so that the energy of each atom is \( \pm \mu \mathrm{H} \). If the magnetic moment \( \mu \) is equal to one Bohr magneton, i.e., \( \mu=0 . \times 10-20 \mathrm{ergs} / \) gauss, below what temperature must one cool the solid so that more than \( 73 \% \) of the atoms are in the lower energy state?A solid at absolute temperature \( T \) is placed in an external magnetic field \( \mathrm{H}=30,000 \) gauss. I have this answer but I need the steps how to get T 6.3 (a) The probability that a spin is parallel to the field is eH/KT P=e"H/KT 11+2uH (1) which yields For \( \mathrm{P}=.75 \) and \( \mathrm{H}=30,000 \) gauss, \( T=3.66^{*} \mathrm{~K} \)
Added by Meredith H.
Close
Step 1
This means that the probability of an atom being in the lower energy state is 0.73. Show more…
Show all steps
Your feedback will help us improve your experience
Prabhu Ramji and 88 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Consider an electron in an external magnetic field of $1.0 \mathrm{~T}$, and recall that the energy of an electron in a magnetic field is given by Eq. (9.18) as $E=m \mu_{\mathrm{B}} B$, where $m=\pm 1 / 2$. (a) Make a plot of the ratio of the probability that the electron's moment is aligned with the field (the low-energy state) to the probability that it is anti-aligned (the high-energy state) versus the absolute temperature $T$. (b) What is the ratio at room temperature? (c) How low does the temperature have to be in order for this ratio to be $2 ?$
Statistical Mechanics
The Boltzmann Factor
$\textbf{Effective Magnetic Field.}$ An electron in a hydrogen atom is in the $2p$ state. In a simple model of the atom, assume that the electron circles the proton in an orbit with radius $r$ equal to the Bohr-model radius for $n$ = 2. Assume that the speed $v$ of the orbiting electron can be calculated by setting $L$ = $mvr$ and taking $L$ to have the quantum-mechanical value for a $2p$ state. In the frame of the electron, the proton orbits with radius $r$ and speed $v$. Model the orbiting proton as a circular current loop, and calculate the magnetic field it produces at the location of the electron. also absorbed when the initial state is still the ground state. What is the value of $n^2$ for the final state in the transition for which this wavelength is absorbed, where $n^2$ = $n_x^2$ + $n_Y^2$ + $n_Z^2$ ? What is the degeneracy of this energy level (including the degeneracy due to electron spin)?
Consider a solid containing $N$ atoms per unit volume, each atom having a magnetic dipole moment $\vec{\mu}$. Suppose the direction of $\vec{\mu}$ can be only parallel or antiparallel to an externally applied magnetic field $\vec{B}$ (this will be the case if $\vec{\mu}$ is due to the spin of a single electron). According to statistical mechanics, the probability of an atom being in a state with energy $U$ is proportional to $e^{-L k Y}$, where $T$ is the temperature and $k$ is Boltzmann's constant. Thus, because energy $U$ is $-\vec{\mu} \cdot \vec{B}$, the fraction of atoms whose dipole moment is parallel to $B$ is proportional to $e^{\mu B i k Y}$ $\vec{B}$ is proportional to $e^{-\mu 8: T}$. (a) Show that the magnitude of the magnetization of this solid is $M=N \mu \tanh (\mu B / k T)$. Here tanh is the hyperbolic tangent function: $\tanh (x)=\left(e^{x}-e^{-x}\right) /\left(e^{x}+e^{-x}\right)$. (b) Show that the result given in (a) reduces to $M=N \mu^{2} B / k T$ for $\mu B \propto k T$. (c) Show that the result of (a) reduces to $M=N_{\mu}$ for $\mu B: k T$. (d) Show that both (b) and (c) agree qualitatively with Fig. 32-14.
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Watch the video solution with this free unlock.
EMAIL
PASSWORD