Problem#1
How many protons and how many neutrons are there in a nucleus of the most common isotope of (a) silicon, 28Si14; (b) rubidium, 85Rb37; (c) thallium, 205Tl81?
Answer:
The pre-subscript is Z, the number of protons. The pre-superscript is the mass number A. A = + Z N, where N is the number of neutrons.
(a) 28Si14 has 14 protons and 14 neutrons.
(b) 85Rb37 has 37 protons and 48 neutrons.
(c) 205Tl81 has 81 protons and 124 neutrons
Problem#2
Hydrogen atoms are placed in an external 1.65-T magnetic field. (a) The protons can make transitions between states where the nuclear spin component is parallel and antiparallel to the field by absorbing or emitting a photon. Which state has lower energy: the state with the nuclear spin component parallel or antiparallel to the field? What are the frequency and wavelength of the photon? In which region of the electromagnetic spectrum does it lie? (b) The electrons can make transitions between states where the electron spin component is parallel and antiparallel to the field by absorbing or emitting a photon. Which state has lower energy: the state with the electron spin component parallel or antiparallel to the field? What are the frequency and wavelength of the photon? In which region of the electromagnetic spectrum does it lie?
Answer:
(a) When the z-component of S (and µ) is parallel to B, U = –│µz│B = –2.7928µnB. When the z-component of S (and µ) is antiparallel to B, U = +│µz│B = –2.7928µnB. The state with the proton spin component parallel to the field lies lower in energy. The energy difference between these two states is
∆E = 2(2.7928µnB)
hf = 2(2.7928µnB)
(6.62 x 10-34 Js)f = 2(2.7928)(5.051 x 10-27 J/T)(1.65 T)
f = 7.03 x 107 Hz = 70.3 MHz
and then
λ = c/f = (2.998 x 108 m/s)/(7.03 x 107 Hz) = 4.26 m
(b) we use eq. potensial energy for a magnetic dipole
U = –µ.B = –µBcosφ
and the of the associated spin magnetic moment turns out to be related to by
µz = –(2.00232)esZ/2m
the state with the z-component of μ parallel to B has lower energy. But, since the charge of the electron is negative, this is the state with the electron spin component antiparallel to B.
That is, for ms = –½ , the state lies lower in energy.
For the ms = +½ state,
U = (2.00232)(e/2m)(ћ/2)B
U = +½(2.00232)(eћ/2m)B = +½(2.00232)µBB
For the ms = –½ state, U = –½(2.00232)µBB. The energy difference between these two states is
∆E = (2.00232)µBB
hf = (2.00232)µBB
(6.62 x 10-34 Js)f = (2.00232)(9.274 x 10-24 J/T)(1.65 T)
f = 4.62 x 1010 Hz = 46.2 GHz
and then
λ = c/f = (2.998 x 108 m/s)/(4.62 x 1010 Hz) = 6.49 mm
Problem#3
Hydrogen atoms are placed in an external magnetic field. The protons can make transitions between states in which the nuclear spin component is parallel and antiparallel to the field by absorbing or emitting a photon. What magnetic-field magnitude is required for this transition to be induced by photons with frequency 22.7 MHz?
Answer:
When the spin component is parallel to the field the interaction energy is U = –μZB. When the
spin component is antiparallel to the field the interaction energy is U = +μzB. The transition energy for a transition between these two states is ΔE = 2μzB, where μz = 2.7928µn. The transition energy is related to the photon frequency by,
ΔE = hf
2μzB = hf
2(2.7928)(5.051 x 10-27 J/T)B = (6.62 x 10-34 Js)(22.7 x 106 Hz)
B = 0.533 T
Problem#4
Neutrons are placed in a magnetic field with magnitude 2.30 T. (a) What is the energy difference between the states with the nuclear spin angular momentum components parallel and antiparallel to the field? Which state is lower in energy: the one with its spin component parallel to the field or the one with its spin component antiparallel to the field? How do your results compare with the energy states for a proton in the same field (see Example 43.2)? (b) The neutrons can make transitions from one of these states to the other by emitting or absorbing a photon with energy equal to the energy difference of the two states. Find the frequency and wavelength of such a photon.
Answer:
│μz│ = 1.9130µn, where µn = 3.15245 x 10-8 eV/T.
(a) ∆E = 2µZB = 2(1.9130)(3.15245 x 10-8 eV/T)(2.30 T) = 2.77 x 10-7 eV
Since μ and S are in opposite directions for a neutron, the antiparallel configuration is lower energy. This result is smaller than but comparable to that found in the example for protons.
(b) f = ∆E/h = (2.77 x 10-7 eV x 1.602 x 10-19 J/eV)/(6.62 x 10-34 Js)
f = 66.9 MHz
and then
λ = c/f = (2.998 x 108 m/s)/(66.9 x 106 Hz) = 4.48 m
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