GR 8677927796770177 | # Login | Register

GR9277 #77
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{77}
Consider a heavy nucleus with spin 1/2. The magnitude of the ratio of the intrinsic magnetic moment of this nucleus to that of an electron is

1. zero, because the nucleus has no intrinsic magnetic moment
2. greater than 1, because the nucleus contains many protons
3. greater than 1, because the nucleus is so much larger in diamter than the electron
4. less than 1, because of the strong interactions among the nucleons in a nucleus
5. less than 1, because the nucleus has a mass much larger than that of the electron

Quantum Mechanics$\Rightarrow$}Gyromagnetic Ratio

The intrinsic magnetic moment is defined in terms of the gyromagnetic ratio and spin as $\vec{\mu}_s = \gamma \vec{S}$, where $\gamma = \frac{eg}{2m}$ (g is the Lande g-factor).

Thus, one sees that the magnetic moment is inversely related to mass.

The ratio of the magnetic moment of a nucleus to that of an electron is $\mu_n/\mu_e = m_e/m_n << 1$, as in choice (E). (One can cancel out the S since ETS is nice enough to have the nucleus have the same spin as the electron.).

Alternate Solutions
 Ning Bao2008-01-30 07:26:00 a common sense answer: presumably the more magnetically "efficient" (e.g. gets the same result for less material, or mass) thing would have a higher intrinsic magnetic moment. Bang for the buck theory wins again.Reply to this comment
alemsalem
2010-09-26 03:52:11
to see how bigger mass results in a smaller magnetic moment (classically) think of it this way:
for a given (spin) angular momentum, the one with the higher mass will have a smaller angular frequency therefore smaller current which means less magnetic moment.
this isn't a complete explanation, but it's good to have a feeling for where the mass came from.
Ning Bao
2008-01-30 07:26:00
a common sense answer: presumably the more magnetically "efficient" (e.g. gets the same result for less material, or mass) thing would have a higher intrinsic magnetic moment. Bang for the buck theory wins again.
 his dudeness2010-09-05 07:02:29 I think this kind of reasoning could potentially get you into trouble. For example, I could say "magnetic moment, in classsical terms, is generally current times area. since a nucleus has a much larger "area" than an electron, it should have a larger magnetic moment!" which is clearly wrong. A better way to think about this problem is to realize that particle magnetic moments are generally proportional to angular momentum, with the constant of proportionality being the "gyromagnetic ratio" (charge over mass). Since nuclei and electrons have the same charge and spin angular momentum, the only difference will be in the masses, so an electron's magnetic moment will be ~2000 times bigger.
rmyers
2006-12-01 13:05:25
Presumably the nucleus would also have a larger charge so the top term would increase as well. However the charge would only end up being a hundred times larger at most, while the mass is many orders of magnitude larger than an electron.
 student20082008-10-16 08:15:53 You're right! Also, g-factors of electron, proton and neutron differ ($g_e\approx 2$,$g_p\approx 5.6$,$g_n\approx -3.8$). But this is not relevant for the ratio, for the same reason.
 NoPhysicist32017-03-23 13:24:06 best remark!

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 17, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}