|
|
1=500 T2 = 250 η 1
1,5 105 400 260
980 . ... 38%.
Q1 = 4,2 = 590 ... η 1
ν=l p1=0,1 T1=300 p2=0,2
T1 2 ω
.1 V1=12 V2=16 . . . h
ν=1 : 1) Q1
2/3 Q1 T2 280 T1
T1 470 T2 280 A =100 .
A1 5 A2 ... η 0,2
ν=1 p1 =250 V1 = 10 T2 = 400
: T2 = 300 T3 = 600
ν = 2 n = 2 2
T1 T2 Q=42 A
T2=290 T1 = 400 η
1 (n=4) ω
γ =1,4 V1 = 0,1 3 V2 = 0,3 3 P2 = 2∙105
ν = 2 2 4
ν = 2 V1 = 10 1 = 250 T2 = 500
= 600 T1 500 2 – 300 : 1) ; 2) ,
ν = 3 V1 = 5 1 = 1 T2 = 500
70% . , 5 : 1)
5,5 1,1 : 1) ; 2)
0,4 400
T1=500 T2=300 2 : 1) ... ;
g h h=0 =0
γ : ) : )
ν = 2 n = 2 n = 2
. . . 0,3 300
λ = 1884
200 n m
t1 = 70 1 = 5∙106 t2 = 140
m = 30 g
S = 300
ω=3/4 T1 k=M/m
50 2 1,5 , 5
200 1/2
1600
0,2 . 100 0,282 . .
0,6 λ=0,589
2 2 15,9 /
2 υ1 = 1
10 40 1000 /
= 1,00 , l = 10,0 b = 0,100 ( ) L
20 0,4
ℓ = 1 I = 10 H
l = 40 I = 10
l=2
k= 150 / m = 8 υ Δx = 4
M m υ0
1 5040 3 2
=4 / S = 8 2
0 = 2040 / , S = 6 2
10 100
m1 = 10 v = 300 / m2 = 200 k = 25 /
1 2 , f' 2 3 f'
R=20 r=10
R = 13
λ = 656,3 Δλ = 2,5
ψn(x) = √(2/l)•sin(
- 11H 42He
1,37•109 3 Δt=1 ° ρ 1,03•10
v = 649 /(∙); = 912 /(∙)
0,48
U = 50cos104πt 0,1
1 25 / 1 ρ1=0,9 /3
n1= 1,528 λ1 = 0,434 n2 = 1,523 λ2 = 0,486
(CsCl)
(NaCl)
N = 100 ( ) R1 = 10,0 R2 = 40,0
= 0,250 n=1000 -1 U : a)
20 10
8 135 27
cv V : ) = T
I0 b (. .) R
L = 0,1 R = 0,8 t = 30
l = 50 d1 =1 Δl =10 d2
4 216 72
l = 20 D = 2 N = 200 s = 1 2
l = 50 S = 10 2 N = 1000 L
8 810
25, 25 50 110 220
E B (E<<B) y z – m
1,0
= 0,5∙10−6
R1 = 5 R2 = 10 Q1 = 2∙10-8 Q2 = –10-8
A1=A2=1
3•10-8
S = 2 2 t = 0,5 p = 3•10-9 •/
p m0c (m0 – ) υ ( )
m0c 1) 2)
L 0,5 l=0,6 D=2
1 20 2 0,4
L 0,2 l = 0,5 D = 1 n
L ( ) 0,1 I W 100
L 0,5 C λ=300
L l=1 1,6 S 20 2 n
L 0,02 ψ , I= 5
0,5 ... 0,1 25 5
L l = 60 S = 4 2 4•10-7 2•10-3
= 1,2 ()
B1=0,5 B2=1 ω
2,3
α
8
I=1 /2 <ω>
Li+ U1 = 400 Na+ — U2 = 300
υ 0,8
(=1,01 ) T ( ) pm 1,6•10−14 •2
1 21 5
= 1 B = 21 m
0,8
m1 = 6,5•10-26 m2 = 6,8•10-26 U = 0,5 = 0,5
=3200
45Ca20 164 1
h = 2 R0 = 6,37∙106
h = 500
dxdp=>h l
ΔxΔp ≥ ħ l ≈ 0,1
Nmax n, l
-
u=υ/υ
ΔΔt ≥ : 1) ; 2) τ
Nmax : 1) n, ℓ, mℓ, ms; 2) n, ℓ, mℓ; 3) n, ℓ; 4) n.
∆λ
10-13
f(ε)
f(ε) = 2/√π(kT)-3/2ε1/2e-ε/kT <ε>
f(u)=4/√πe-u2u2 (u=υ/ υ) ΔN υ 0,002
f(υ) = 4π(m0/2πkT)3/2υ2e-m0υ2/2kT <υ>
– – I A R
– – R = 10 I 5
: 1) ; 2)
1 = 0,2cos(2πt + π/12) 2 = 0,2cos(2πt + π/6)
rν,T = 2πhν3/c21/(ehν/kT – 1) hν << k –
rν,T = 2πhν3/c21/(ehν/kT – 1) –
rν,T = 2πhν3/c21/(ehν/kT – 1)
ΔRe Δλ = 1
- 60 / 0,25
Δλ = 5
1) 2)
ψ = A•–r/a r –
ψ(r) = A•e–r2/(22) r —
n = 2 λ = 1,212∙10
Δυ 1
1
Δ Δ
l Emin=10
(m = 6,68•10–27 ) = 3,29
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