PHL424: Nuclear angular

orbitals in an

quantum numbers: n=3 n (principal) 1,2,3,… ℓ (orbital ) 0 → n-1 m (magnetic) -ℓ ≤ m ≤ +ℓ n=2 s () ↑↓ or +½ħ -½ħ n=1

classical analogy

spin s

orbital angular momentum ℓ

sun ≡ nucleus earth ≡ electron and have ℓ and s electron is structure less and hence can not rotate

spin s is a quantum mechanical concept total angular momentum: = +

total nuclear spin: = 𝚥𝚥⃗ ℓ 𝑠𝑠⃗ 𝐼𝐼 ∑ 𝑗𝑗 Indian Institute of Technology Ropar Hans-Jürgen Wollersheim - 2017 Nuclear spin

protons and neutrons have orbital angular momentum ℓ and spin s

total angular momentum: = +

total nuclear spin: = 𝚥𝚥⃗ ℓ 𝑠𝑠⃗ = + + , + + 1, , 𝐼𝐼 ∑ 𝑗𝑗 𝐼𝐼 𝑗𝑗1 𝑗𝑗2 ⋯ 𝑗𝑗𝑛𝑛 𝑗𝑗1 𝑗𝑗2 ⋯ 𝑗𝑗𝑛𝑛 − ⋯ 𝑗𝑗1 − 𝑗𝑗2 − ⋯ − 𝑗𝑗𝑛𝑛

Indian Institute of Technology Ropar Hans-Jürgen Wollersheim - 2017 Nuclear spin quantum number

protons and neutrons have orbital angular momentum ℓ and spin s

total angular momentum: = +

total nuclear spin: = 𝚥𝚥⃗ ℓ 𝑠𝑠⃗ = + + , + + 1, , 𝐼𝐼 ∑ 𝑗𝑗 quantum mechanics 𝐼𝐼 𝑗𝑗1 𝑗𝑗2 ⋯ 𝑗𝑗𝑛𝑛 𝑗𝑗1 𝑗𝑗2 ⋯ 𝑗𝑗𝑛𝑛 − ⋯ 𝑗𝑗1 − 𝑗𝑗2 − ⋯ − 𝑗𝑗𝑛𝑛 . 1H = 1 , so I = ½ . 2H = 1 proton and 1 , so I = 1 or 0

. For larger nuclei, it is not immediately evident what the spin should be as there are a multitude of possible values.

mass number number of protons number of neutrons spin (I) example

even even even 0 16O

odd odd integer (1,2,…) 2H

odd even odd half-integer ( , , ) 13C 1 3 2 2 odd even half-integer ( ⁄ , ⁄ , ⋯) 15N 1 3 ⁄2 ⁄2 ⋯

Indian Institute of Technology Ropar Hans-Jürgen Wollersheim - 2017 Nuclear spin quantum number

The magnitude is given by

= + 1

The projection on the z-axis (arbitrarily chosen), takes on discretized values according to m, where 𝐿𝐿 ℏ 𝐼𝐼 𝐼𝐼 = , + 1, + 2, , +

𝑚𝑚 −𝐼𝐼 −𝐼𝐼 −𝐼𝐼 ⋯ 𝐼𝐼

Indian Institute of Technology Ropar Hans-Jürgen Wollersheim - 2017 Parity

wave – particle duality: photoelectric effect

wave function

Ψ(x)

Ψ(x) = Ψ(-x) → parity = even (+) x Ψ(x) ℓ = 0, 2, 4, … even

x Ψ(x) = -Ψ(-x) → parity = odd (-)

ℓ = 1, 3, 5, … odd

Indian Institute of Technology Ropar Hans-Jürgen Wollersheim - 2017

=

𝜇𝜇⃗ 𝐼𝐼 ∙ 𝐴𝐴 A

2 = = = 𝜋𝜋𝜋𝜋 2𝜋𝜋𝜋𝜋 ℓ 𝑚𝑚𝑒𝑒𝑣𝑣 ∙ 𝑟𝑟 𝑣𝑣 → 𝑡𝑡 = = 𝑡𝑡 𝑣𝑣 2 𝑒𝑒 𝑒𝑒 ∙ 𝑣𝑣 I 𝐼𝐼 = 𝑡𝑡 𝜋𝜋𝜋𝜋 = 2 𝑒𝑒 ∙ 𝑣𝑣 2 𝑒𝑒 ∙ 𝑣𝑣 ∙ 𝑟𝑟 ℓ 𝜇𝜇 ∙ 𝜋𝜋𝑟𝑟 = 2𝜋𝜋𝜋𝜋 = = 2 2 2 𝑒𝑒 ∙ 𝑣𝑣 ∙ 𝑟𝑟 𝑚𝑚𝑒𝑒𝑣𝑣 ∙ 𝑟𝑟 𝑒𝑒 ∙ ℓ 𝑒𝑒 ∙ ℏ 𝜇𝜇 ∙ 𝜇𝜇𝐵𝐵𝐵𝐵퐵𝐵𝐵 𝑚𝑚𝑒𝑒𝑣𝑣 ∙ 𝑟𝑟 𝑚𝑚𝑒𝑒 𝑚𝑚𝑒𝑒 = electron orbital magnetic moment ℓ 𝜇𝜇ℓ 𝜇𝜇𝐵𝐵 ∙ = ℏ proton orbital magnetic moment = 2 ℓ 𝑒𝑒 ∙ ℏ 𝜇𝜇ℓ 𝜇𝜇𝑁𝑁 ∙ 𝜇𝜇𝑁𝑁 Indian Institute of Technology Ropar Hans-Jürgen Wollersheimℏ - 2017 𝑚𝑚𝑝𝑝 Magnetic moment

= electron orbital magnetic moment ℓ 𝜇𝜇ℓ 𝜇𝜇𝐵𝐵 ∙ ℏ = 2.0023 electron () 𝑠𝑠 𝜇𝜇𝑠𝑠 − ∙ 𝜇𝜇𝐵𝐵 ∙ ℏ = +5.585691 proton spin magnetic moment 𝑠𝑠 𝜇𝜇𝑠𝑠 ∙ 𝜇𝜇𝑁𝑁 ∙ ℏ = 3.826084 neutron spin magnetic moment 𝑠𝑠 Why has a neutron a magnetic moment when it is uncharged? 𝜇𝜇𝑠𝑠 − ∙ 𝜇𝜇𝑁𝑁 ∙ ℏ

+ u-quark: neutrons and protons are not elementary particles d-quark: 2 3 internal structure: they have charges. 1⁄ 𝑒𝑒 3 proton neutron − ⁄ 𝑒𝑒 +1e 0e

1 = 5.59 3.83 = 0.87980 = 0.8574 (experiment) 2 2 𝜇𝜇𝑠𝑠 1𝐻𝐻 − ∙ 𝜇𝜇𝑁𝑁∙ 𝜇𝜇𝑁𝑁 ∙ 𝜇𝜇𝑁𝑁

Indian Institute of Technology Ropar Hans-Jürgen Wollersheim - 2017 Magnetic resonance imaging (MRI)

= +5.585691 proton spin magnetic moment 𝑠𝑠 𝜇𝜇𝑠𝑠 ∙ 𝜇𝜇𝑁𝑁 ∙ = ℏ gyromagnetic ratio = = 47.89 10 𝜇𝜇𝑁𝑁 6 −1 −1 𝜇𝜇𝐼𝐼 𝛾𝛾 ∙ 𝐼𝐼 proton -factor: +5.585691,𝛾𝛾 𝑔𝑔 ∙ ℏspin 𝑔𝑔I: ∙½ ħ ∙ 𝑇𝑇 𝑠𝑠 𝑔𝑔 proton in energy difference between states =

∆𝐸𝐸 = ℎ2 ∙ 𝜈𝜈

Δ𝐸𝐸= ∙ 𝜇𝜇𝐼𝐼 ∙ 𝐵𝐵 0 Larmor frequency 𝛾𝛾 𝜈𝜈 =⁄2𝜋𝜋42∙.𝐵𝐵570 for proton 𝛾𝛾 𝑀𝑀𝑀𝑀𝑀𝑀 ⁄2𝜋𝜋 ⁄𝑇𝑇

Larmor frequency low energy

high energy

low energy

high energy

Indian Institute of Technology Ropar Hans-Jürgen Wollersheim - 2017