Solution to all your problems:
1:
B = 300 gauss = 0.03 T
E= 200/0.002 V/m
Vdrift = (E x B)/B2 ≈ 0,33 x E7 m/s
2.
At start position (“0”): v0perp = vtot * sinθ, B = B0, (v0tot)2 = (v0perp)2 + (v0parallell)2
At reflection: vperp = vtot, , B = Br
½ m v2perp/B is a preserved quantity =>
 (v0perp)2/B0 = (vtot*sinθ)2/B0 = (vperp)2/BR = (vtot)2/BR
 Br = B0 / (sinθ)2
If B never reaches this condition (sinθ too small) then vperp never reaches zero and
there is a remaining non-zero vparallell => the particle is lost from the trap.
3.
p = 2mbar = 2 hP = 200P. pV = nkT, ne = 0.01*p/kT ≈ 3,6 * 1020 m-3
ω2p = (e2 * ne)/(2*ε0*m) => ωp ≈ 760 GHz, λ = c*2*π/ω ≈ 2,5 mm
4.
vrms = SQRT(3RT/M) e: vrms = 6,7*106 m/s p: vrms = 1,6*105 m/s
5.
ions: E= N* 3/2 kT = n*V*3/2 kT = 240J, Etot = 480J, 1cal = 4,19J, m=200g => ∆T = 0,6K
7.
See Spectrophysics 9.6.1 and the discussion therein.
9.
See Spectrophysics 10.1 .
10.
In LS-coupling (i.e. for most light elements), the selection rule ∆s = 0 is valid.
11.
See Spectrophysics section 9.4 .
12.
See Spectrophysics section 10.2 .
13.
1
S0 :
S1/2 :
1
P1 :
3
P2 :
3
F4 :
5
D1 :
1
D2 :
6
F9/2 :
2
14.
S
0
½
0
1
1
2
½
5/2
L
0
0
1
1
3
2
2
3
J
0
1/2
1
2
4
1
2
9/2
n3/n2 = g3/g2 * exp(-∆E/kT)
n3/n2 = 0,01, n = 2: 2s1/2 2p1/2 2p3/2 . Stat w. = 2J+1 gives g2 = 2+2+4 = 8
n = 3: 3s1/2 3p1/2 3p3/2 3d3/2 3d5/2 Stat w. = 2J+1 gives g3 = 2+2+4+4+6 = 18
∆E = 13.6*(1/32 – 1/22) eV ≈ -1.89 eV. Note: E = 0 for the ground state in section 9.5.1
in Spectrophysics! (Theoretical) atomic physicists set E=0 at the ionization limit!
Beware!
kT = ln((n3*g2)/(n2*g3))/∆E => T ≈ 4000 K.
15.
See 3. Above! ν ≈ 0,2 GHz
16.
E: 3/2 kT, vrms = SQRT(3RT/M) = SQRT(3kT/m)
17.
a: 1̴ 00%, b: 1̴ %, c: 1̴ 00%
18.
IP = 10,5eV, ntot = 1,25 x 1021 m-3 , Qi/Qa = 1
kT = 2eV:
100% joniserat
kT = 1eV:
10% joniserat
kT = 0.5eV: 3.5% joniserat
19.
Te : 1eV < - > Ti: 40oC . See handout (Waymouth).
20.
See Spectrophysics section 9.4 .
21.
The selection rule ∆J = 0 or ± 1 but 0 -> 0 forbidden is valid also when In LS-coupling
breaks down.