Chapter 1 - Stern Gerlach Experiments
Problem 1.2
Show for a solid spherical ball of mass $m$ rotating about an axis through its center with a charge $q$ uniformly distributed on the surface of the ball that the magnetic moment $\mu$ is related to the angular momentum $L$ by the relation
$$\mu = \frac{5q}{6mc}L$$
Solution
We are going to work in Gaussian units.
Recall that the magnetic moment is given by $$\mu = \frac{IA}{c}$$. Since we are dealing with uniformly charged sphere, we will cut up the sphere into infinitely many rings that are infinitely small in height. Without loss of generality we can cut them horizontally so that they are parallel to the x-axis.
Because the ball is rotating around the y-axis, there is an induced current. The current for each ring is given by $$I = \frac{\sigma dA}{T}$$, where $\sigma = \frac{q}{4\pi R^2}$ is the surface charge density for a sphere of radius $R$ and $T = \frac{2\pi}{\omega}$ is the period of the sphere's rotation.
Assuming the sphere has radius R, each ring has radius $r = Rsin\theta$ and thus $dA = 2\pi r Rd\theta$ is the area of each ring.
Thus $d\mu = \frac{dIA}{c} = \frac{dI}{c}\pi R^2sin^2\theta$.
Then to obtain $\mu$ of the whole sphere we need to integrate over all angles to get $$\mu = \int d\mu = \int \frac{\sigma dA}{Tc}\pi R^2sin^2\theta = \int\limits_0^{\pi} \frac{\sigma}{Tc}2{\pi}^2 R^4sin^3\theta d\theta \\ = \int\limits_0^{\pi} \frac{\pi q}{2Tc} R^2sin^3\theta d\theta \\ = \frac{q}{3c}\frac{2\pi}{T}R^2 \\ = \frac{q}{3c}\omega R^2 $$
Since the angular momentum of sphere is given by $$L = \frac{2}{5}mR^2\omega$$ we finally get $$\mu = \frac{5q}{6mc}L$$ as required.
Showing posts with label Townsend. Show all posts
Showing posts with label Townsend. Show all posts
Monday, May 12, 2014
Wednesday, May 7, 2014
Townsend - Problem 1.1
Chapter 1 - Stern Gerlach Experiments
Problem 1.1
Determine the field gradient of a 50-cm-long Stern Gerlach magnet that would produce a 1-mm separation at the detector between spin-up and spin-down silver atoms that are emitted from an oven at $T = 1500 K$. Assume the detector is located 50 cm from the magnet. Note that the emitted atoms have average kinetic energy $2k_bT$.
Solution
We are interested in the field gradient necessary to produce a given pattern on the detector, hence we need to find $\frac{\partial B}{\partial z}$.
From equation 1.4 we have that the force due to a Stern-Gerlach magnet is given by
$$F_z = \mu_z \frac{\partial B}{\partial z}$$,
where $\mu_z$ is the magnetic dipole moment of the particle given by
$$\mu_z = \frac{gq}{2mc}S_z$$ with $S_z$ being the intrinsic spin angular momentum of the particle in the z-direction.
For silver atoms, the magnetic dipole moment is due to the intrinsic spin of a single electron. Hence $g = 2$ and $q = e$, the charge of an electron.
Notice that, using symmetry arguments, the displacement in the z-direction of the upper particle is the same as the displacement of the lower particle. Let this displacement be $z$. Then $d = 1 mm = 2z$.
Therefore, we can focus on the displacement in the z-direction of the upper particle.
While the particle is traveling through the magnet it experiences a force and thus accelerates in the z-direction. We have
$$a_z = \frac{F_z}/{M} = \frac{\mu_z}/{M} \frac{\partial B}{\partial z}$$ where $M$ is the mass of the silver atom.
When the magnet is not traveling through the magnet it does not accelerate in the z-direction.
Initially there is no velocity in the z-direction and we have that the displacement in the z-direction of the upper particle while it is traveling through the magnet is given by
$$z_1 = \frac{1}{2}a_zt^2 = \frac{1}{2}a_z \left(\frac{l}{v_x}\right)^2$$, where $v_x$ is the initial velocity of the particle in the x-direction.
Outside the magnet the displacement in the z-direction of the upper particle is given by:
$$z_2 = v_zt = v_z\frac{l}{v_x} = a_z\left(\frac{l}{v_x}\right)^2$$
because v_z is the velocity of the particle when it just left the magnet.
The total displacement is thus
$$z = z_1 + z_2 = \frac{3}{2}a_z\left(\frac{l}{v_x}\right)^2$$
We can solve for $v_x$ by making use of the fact that the particles are ejected with average kinetic energy $2k_bT$.
$$\frac{1}{2}M{v_x}^2 = 2k_bT$$
Thus ${v_x}^2 = \frac{4k_bT}{M}$ and we can solve
$$d = 2z = 3a_zl^2\frac{M}{4k_bT} = \frac{3}{4}\frac{\mu_z}{M} \frac{\partial B}{\partial z}\frac{4k_bTl^2}{M}$$
Finally, we rearrange for $\frac{\partial B}{\partial z}$ and plug in the values:
$$\frac{\partial B}{\partial z} = 1200 G/cm$$
Problem 1.1
Determine the field gradient of a 50-cm-long Stern Gerlach magnet that would produce a 1-mm separation at the detector between spin-up and spin-down silver atoms that are emitted from an oven at $T = 1500 K$. Assume the detector is located 50 cm from the magnet. Note that the emitted atoms have average kinetic energy $2k_bT$.
Solution
We are interested in the field gradient necessary to produce a given pattern on the detector, hence we need to find $\frac{\partial B}{\partial z}$.
From equation 1.4 we have that the force due to a Stern-Gerlach magnet is given by
$$F_z = \mu_z \frac{\partial B}{\partial z}$$,
where $\mu_z$ is the magnetic dipole moment of the particle given by
$$\mu_z = \frac{gq}{2mc}S_z$$ with $S_z$ being the intrinsic spin angular momentum of the particle in the z-direction.
For silver atoms, the magnetic dipole moment is due to the intrinsic spin of a single electron. Hence $g = 2$ and $q = e$, the charge of an electron.
Notice that, using symmetry arguments, the displacement in the z-direction of the upper particle is the same as the displacement of the lower particle. Let this displacement be $z$. Then $d = 1 mm = 2z$.
Therefore, we can focus on the displacement in the z-direction of the upper particle.
While the particle is traveling through the magnet it experiences a force and thus accelerates in the z-direction. We have
$$a_z = \frac{F_z}/{M} = \frac{\mu_z}/{M} \frac{\partial B}{\partial z}$$ where $M$ is the mass of the silver atom.
When the magnet is not traveling through the magnet it does not accelerate in the z-direction.
Initially there is no velocity in the z-direction and we have that the displacement in the z-direction of the upper particle while it is traveling through the magnet is given by
$$z_1 = \frac{1}{2}a_zt^2 = \frac{1}{2}a_z \left(\frac{l}{v_x}\right)^2$$, where $v_x$ is the initial velocity of the particle in the x-direction.
Outside the magnet the displacement in the z-direction of the upper particle is given by:
$$z_2 = v_zt = v_z\frac{l}{v_x} = a_z\left(\frac{l}{v_x}\right)^2$$
because v_z is the velocity of the particle when it just left the magnet.
The total displacement is thus
$$z = z_1 + z_2 = \frac{3}{2}a_z\left(\frac{l}{v_x}\right)^2$$
We can solve for $v_x$ by making use of the fact that the particles are ejected with average kinetic energy $2k_bT$.
$$\frac{1}{2}M{v_x}^2 = 2k_bT$$
Thus ${v_x}^2 = \frac{4k_bT}{M}$ and we can solve
$$d = 2z = 3a_zl^2\frac{M}{4k_bT} = \frac{3}{4}\frac{\mu_z}{M} \frac{\partial B}{\partial z}\frac{4k_bTl^2}{M}$$
Finally, we rearrange for $\frac{\partial B}{\partial z}$ and plug in the values:
- d = 0.1 cm
- l = 50 cm
- T = 1500
$$\frac{\partial B}{\partial z} = 1200 G/cm$$
Townsend - A Modern Approach to Quantum Mechanics
Here is an index of my solutions to problems in Townsend's "A Modern
Approach to Quantum Mechanics", 2nd edition. Please note that some
solutions might be incorrect. If you do spot an error, let me know about it
in the comments.
Chapter 1 - Stern Gerlach Experiments
1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15
Chapter 1 - Stern Gerlach Experiments
1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15
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