Parts a), b), c), d), and e). Please show and explain all steps. Some relevant equations are included. Thank you!
Posted: Mon May 09, 2022 2:04 pm
Parts a), b), c), d), and e). Please show and explain
all steps. Some relevant equations are included. Thank
you!
13–9 If the magnetic moment u = gub of an atom is large enough, there will be 2J + 1 possible angles e between the magnetic moment and the applied magnetic intensity H corresponding to magnetic levels having energies ej = MJMH where my has values between -J and +J. (a) Show that Zye will be given by (2J + 1) sinh 2kT Z = (13–60) sinh 2kT MH
[Hint: See the derivation of Eq. (12-44).] (b) Show that the net magnetic moment of the system is given by (2J + 1) M = Nu coth (2J + 1) coth 2 2KT 2kT MH UH ] This is called the Brillouin* function. (c) Show that the net magnetic moment follows Curie's law in the limit of high temperatures and low fields. (d) In the limit of low tem- perature and high fields, show that all the dipoles are aligned. (e) Show that the expression for the net magnetic moment derived in part (b) reduces to Eq. (13-29) when 2J + 1 = 2 and g = 2.
where n; = 0,1,2, ..., and h is Planck's constant. An unexpected result is that the oscillator can never be in a state of zero energy, but that in the lowest level the energy is hv/2, in the next level it is 3hv/2, and so on. The levels are nondegenerate; there is only one energy state in each level; and gs = l in each level. The quantum condition that the energy can have only some one of the set of values [(n; + 1/2)hv] is equivalent to the condition that the amplitude can have only some one of the set of values such that *. - (», + 2) VI/Km. = n+2 + Using Eq. (12-43), the partition function of the assembly is Z = z = {cxp (**) – Şexp [-(n + ) *) , + To evaluate the sum, let z = hv/kt for brevity. Writing out the first few terms, we have Z = exp 3z 2 + +... +(-3) + exp ( - ) + exp(-3) (-3)1 2 = exp - {1 + exp(-2) + [exp (-2)] + ...). The sum in the preceding equation has the form of the infinite geometric series 1 + p + p + ..., which equals 1/(1 - p) as is readily verified by expanding the product (1 – p) x (1 + p + p + ...). Therefore 1 Z = exp -x' exp(-hv/2kT) (12–44) 1 - exp(-hv/kT) xp(-2). - exp(-3) ог Z =
13-4 13-4 PARAMAGNETISM 401 and MBH e The net magnetic moment M of the crystal is the product of the magnetic moment Mg of each ion and the excess number of ions aligned parallel to the field. Then M = - (Ñ1 - Ñub = Nug tanh (13–29) KT This is the magnetic equation of state of the crystal, expressing the magnetic moment M as a function of H and T. Note that M depends only on the ratio H|T. The equation of state can also be derived as follows. The function F* is F* = -Nkt In Z = - NKT In 2 cosh (13–30) kТ The magnetic moment M, which in this case corresponds to the extensive variable X, is aF* (13-31) kТ MBH Nup tanh
all steps. Some relevant equations are included. Thank
you!
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[Hint: See the derivation of Eq. (12-44).] (b) Show that the net magnetic moment of the system is given by (2J + 1) M = Nu coth (2J + 1) coth 2 2KT 2kT MH UH ] This is called the Brillouin* function. (c) Show that the net magnetic moment follows Curie's law in the limit of high temperatures and low fields. (d) In the limit of low tem- perature and high fields, show that all the dipoles are aligned. (e) Show that the expression for the net magnetic moment derived in part (b) reduces to Eq. (13-29) when 2J + 1 = 2 and g = 2.
where n; = 0,1,2, ..., and h is Planck's constant. An unexpected result is that the oscillator can never be in a state of zero energy, but that in the lowest level the energy is hv/2, in the next level it is 3hv/2, and so on. The levels are nondegenerate; there is only one energy state in each level; and gs = l in each level. The quantum condition that the energy can have only some one of the set of values [(n; + 1/2)hv] is equivalent to the condition that the amplitude can have only some one of the set of values such that *. - (», + 2) VI/Km. = n+2 + Using Eq. (12-43), the partition function of the assembly is Z = z = {cxp (**) – Şexp [-(n + ) *) , + To evaluate the sum, let z = hv/kt for brevity. Writing out the first few terms, we have Z = exp 3z 2 + +... +(-3) + exp ( - ) + exp(-3) (-3)1 2 = exp - {1 + exp(-2) + [exp (-2)] + ...). The sum in the preceding equation has the form of the infinite geometric series 1 + p + p + ..., which equals 1/(1 - p) as is readily verified by expanding the product (1 – p) x (1 + p + p + ...). Therefore 1 Z = exp -x' exp(-hv/2kT) (12–44) 1 - exp(-hv/kT) xp(-2). - exp(-3) ог Z =
13-4 13-4 PARAMAGNETISM 401 and MBH e The net magnetic moment M of the crystal is the product of the magnetic moment Mg of each ion and the excess number of ions aligned parallel to the field. Then M = - (Ñ1 - Ñub = Nug tanh (13–29) KT This is the magnetic equation of state of the crystal, expressing the magnetic moment M as a function of H and T. Note that M depends only on the ratio H|T. The equation of state can also be derived as follows. The function F* is F* = -Nkt In Z = - NKT In 2 cosh (13–30) kТ The magnetic moment M, which in this case corresponds to the extensive variable X, is aF* (13-31) kТ MBH Nup tanh