Content uploaded by Andrea Rossi
Author content
All content in this area was uploaded by Andrea Rossi on Dec 19, 2021
Content may be subject to copyright.
Au197 +N14 →Au198 +N13
σI = 2
N
RN
RN≈~√N
2mec=c√N
2ωe
=re√N
2,
re=c
ωe=λe
2π
ωe=mec2
/~
ρ(ω) :
ρ(ω) = ~ω3
2π2c3dω.
N= 1011 D
D= 2RN≈0.12 µm
dE
dE≈r4πR2
N
N=√πre≈1.78re≈0.68 ·10−12 m.
N
re=λe
/2π≈0.38 ·10−12 m.
FCA
FC(d)
A=π2~c
240d4.
d c
A=π(λe
/4π)2
FC(d)
Fe(d) :
FC(d) = π~cλ2
e
3840d4,
Fe(d) = 1
4πε0
e2
d2.
db≈2λe
/2π≈0.77 ·10−12m
re
A=π(λe
/2π)2=πr2
e
db≈4λe
/2π≈
1.54 ·10−12m
ΦM=h
/e
h e
e c
FL
FL(d) = ecB (d) = µo
4π·e2c2
d2=1
4π0·e2
d2,
B(d) = µoec
4πd2
c~
d
d
d=nλe
±~
/2~
~
FL
LD
LD=
N
X
a=1 hma
2v2
a−ea
2φa(ra) + ea
2cva·Aa(ra)i
Aa(ra) =
N
X
b6=a
eb[vb+ (vb·ruab)ruab ]
2crab
φa(ra) =
N
X
b6=a
eb
rab
rab =|ra−rb|
ruab =ra−rb
|ra−rb|
a b
raeavamaAa(ra)
φ(ra)raN
pa
pa=mava=ea
cAazp
Aazp Aaz
va
ma
2v2
a=p2
a
2ma
p2
a
2ma
=e2
aA2
azp
2c2ma
ma=eaAaz
c2
p2
a
2ma
=eaA2
azp
2Aaz
Aazp
Aaz 'va
c
vaAazp
ma
2v2
a=eavaAazp
2c=ea
2cva·Aazp
Aat (ra) = Aazp +Aa(ra)
LD=
N
X
a=1 h−ea
2φa(ra) + ea
2cva·Aat (ra)i
Aazp
Aaz
Aaz.
e
~=c= 1,
Lz=
N
X
a=1
[eaca·Aa(ra)−eaφa(ra)]
L=T−U
T=
N
X
a=1
eaca·Aa(ra)
U=
N
X
a=1
eaφa(ra)
dϕaM =eaAa(ra)·dl
dl=cadt
dϕaM =eaAa(ra)·cadt
dϕaE =eaφa(ra)dt
dt
ea, a
ωzbw =dϕaM
dt =ma
T=
N
X
a=1
ma
Aa(ra)eaca
αrea
Aa(ra) = eaca
αrea
+X
b6=a
eb[cb+ (cb·ruab)ruab ]
rab
φa(ra) = ea
αrea
+X
b6=a
eb
rab
Lz=
N
X
a=1 (1
rea
+X
b6=a
αca·cb+α(cb·ruab) (ca·ruab )
rab −"1
rea
+X
b6=a
α
rab #)
Lz=
N
X
a=1 X
b6=a
α[ca·cb+ (cb·ruab) (ca·ruab )−1]
rab
rab =ra−rb
rab =|ra−rb|
ruab =rab
rab
ca·cb= cos (ϑab1)
ca·ruab = cos (ϑab2)
cb·ruab = cos (ϑab3)
ϑab1=ϑab2−ϑab3
Lz=
N
X
a=1 X
b6=a
α[cos (ϑab2−ϑab3) + cos (ϑab2) cos (ϑab3)−1]
rab
Lzab =α[cos (ϑab2−ϑab3) + cos (ϑab2) cos (ϑab3)−1]
rab
raeaca
(c2
a= 1) α=e2
a(α−1≈137.036) rea
rab tab
ruab rab
(mea =r−1
ea )αrea
δ(S)=0
S=ˆ4T
Lzdt.
S=
N
X
a=1 X
b6=aˆ4T
α[cos (ϑab2−ϑab3) + cos (ϑab2) cos (ϑab3)−1]
rab
dt
S=
N
X
a=1 X
b6=aˆ4T
Lzabdt
δ(Lzab) = 0 =⇒δ(S) = 0
δ(Lzab (rab, ϑab2, ϑab3)) = ∂Lz
∂rab
δrab +∂Lz
∂ϑab2
δϑab2+∂Lz
∂ϑab3
δϑab3= 0
∂Lz
∂rab
=−α[cos (ϑab2−ϑab3) + cos (ϑab2) cos (ϑab3)−1]
r2
ab
∂Lz
∂ϑab2
=α[−sin (ϑab2−ϑab3)−sin (ϑab2) cos (ϑab3)]
rab
∂Lz
∂ϑab3
=α[sin (ϑab2−ϑab3)−cos (ϑab2) sin (ϑab3)]
rab
r2
ab −c2t2
ab = 0
(ϑab1= 2πn)∩ϑab2=π
2+πm=⇒δ(S) = 0 (n, m ∈Z).
(tz'8.1·10−21s)
ωzbw
m
ωzbw =m
m m0Ek
m=m0+Ek
γt
S
A
A=A+γtφ
·A=S
γt,
U
P
1
8πAγtg
A=Uγt+P
P=−1
4π(E×B− SE)
P
S
E.
(·A= 0) SE
SE
/4π
P
ρ
SρS
4πρ =∇ · E
ρS=1
4π∇ · ES
dU
dt =ρS
φ
dU
dt =ρdφ
dt
eS =edφ
dt
δωzbw
ωzbw fS
eφ =dϕ
dt
δωzbw =dϕ
dt
eS=deφ
dt
eS=d2ϕ
dt2
eS=dωzbw
dt
φ=ˆSdt
fS=−e∇φ
eS.
S.
(λe≈2.43 ·10−12m)
I= 0.25A
dne
dt =I
e= 1.56 ·1018ne
/s
Ee=1
4πε0
e2
λe
wout =Ee
dne
dt '150w
(10−12 m) (10−10 m)
(10−15 m)
rq=~
/2mec≈0.192
re
(rq=re
/2)
~
re= 0.38 pm Bzbw
Bzbw = 32.21 ·106T.
gH
gH= 267.52 ·106rad ·s−1·T−1
νNMR =gHBz bw
2π= 1.3714 ·1015 Hz
νp
νp=νNMR
/2= 6.8571 ·1014 Hz.
λp=c
νp
= 4.372 ·10−7m
(gD= 41.066 ·106rad ·s−1·T−1)
220 V
(200 kV )
b
λmax
Tk=b
λmax
Tk=2.898 ·10−3
0.3575 ·10−6= 8106 K.
Wout =σεT 4
kA≈22 kW
Eout = 22 k W h
σ= 5.67 ·10−8W m−2K−4ε= 0.9
A≈10−4m2l≈1cm
d≈0.3cm
m3
/h'kg
/h
Kcal
/hkWh
/h
m3
/h
CO2
Einp
Einp = 380 W h
COP =Eout
Einp ≈54
R.P.Y∞X
σI=2