פתרון בגרות · קרינה וחומר

✦ בנוי על ידי פהד גאנם ✦

 ·  AS₁ = 2λ  ;  AS₂ = 5λ  ;  BS₁ = BS₂
A₀ = 0.2 cm  ;  f = 12.5 Hz  ;  λ = vf  ;  T = 1f
 ·  X = 1f [s]
X  |  0.050  |  0.059  |  0.071  |  0.100  |  0.143
ΔL [cm]  |  1.52  |  1.78  |  2.13  |  2.98  |  4.21
12.5 Hz
3
2.00 λ
5.00 λ
2.40 cm
3.00
n = 3
0.80 cm
0.080 s
0.40 cm

AS₁ = 2λ   ;   AS₂ = 5λ
ΔL = AS₂ − AS₁ = 5λ − 2λ = 3λ
n·λ = 3λ   ⟹   n = 3

ΔL = n·λ   ;   n = 0, 1, 2, 3, …
v = λ·f   ⟹   λ = vf
ΔL = n·λ = n·vf = n·vf
ΔL = 3vf = 3v · 1f

a = ΔL₂ − ΔL₁X₂ − X₁ = 4.21 − 1.520.143 − 0.050 = 2.690.093 ≈ 28.9 cm/s a ≈ 30 cm/s
a = n·v = 3v   ⟹   v = an = 303 = 10 cm/s
2.980.100 = 29.8 cm/s  ⟹  v = 29.83 ≈ 10 cm/s

BS₁ = BS₂   ⟹   ΔL = BS₂ − BS₁ = 0
ΔL = n·λ  ;  0 = n·λ  ⟹  n = 0
A_B = A₀ + A₀ = 2A₀ = 0.2 + 0.2 = 0.4 cm

A_B = A₁ + A₂ = 0.2 + 0.2 = 0.4 cm
T = 1f = 112.5 = 0.08 s
y(0) = −A_B = −0.4 cm y(t) = −A_B·cos(2πtT) = −0.4·cos(2πt0.08)
T4 = 0.084 = 0.02 s y(0) = −0.4  ;  y(0.02) = 0  ;  y(0.04) = +0.4  ;  y(0.06) = 0  ;  y(0.08) = −0.4  [cm]
λ = vf = 1012.5 = 0.8 cm ΔL(A) = 3λ = 3 · 0.8 = 2.4 cm

A: ΔL = 3λ = n·λ  ⟹  n = 3 B: ΔL = 0 = 0·λ  ⟹  n = 0
A_B = A₁ − A₂ = 0.2 − 0.2 = 0 cm

L = 0.75 m  ;  d = w  ;  sinθ₁ = λw  ;  Δx = 2·λ·Lw

w [mm]  |  0.016  |  0.020  |  0.030  |  0.040  |  0.080
Δx [m]  |  0.040  |  0.035  |  0.025  |  0.017  |  0.010
1w [1/mm]  |  62.5  |  50.0  |  33.3  |  25.0  |  12.5
0.030 mm
0.75 m
400 nm
0.010 m
0.020 m
2.0 cm
0.010 m
0.0133
0.76°
33.3 1/mm
6.0·10⁻⁴ m·mm
6.0·10⁻⁷ m²
400 nm
4·10⁻⁴ m
0.030 mm
1 + 2N

sinθ_m = m·λw  ;  m = 1, 2, 3, … sinθ₁ = λw
tanθ₁ = y₁L ≈ sinθ₁ = λw y₁ = λ·Lw
Δx = 2·y₁ = 2·λ·Lw

Δx = 2·λ·Lw = (2·λ·L)·1w
1 mm = 10⁻³ m  ⟹  1 1/mm = 10³ · 1/m
10.016 = 62.5  ;  10.020 = 50.0  ;  10.030 = 33.3  [1/mm] 10.040 = 25.0  ;  10.080 = 12.5  [1/mm]

(12.5 , 0.010)  ;  (25.0 , 0.017)  ;  (33.3 , 0.025)  ;  (50.0 , 0.035)  ;  (62.5 , 0.040)

a = 2·λ·L   ⟹   λ = a2·L
a = Δx₂ − Δx₁X₂ − X₁ = 0.040 − 0.01062.5 − 12.5 = 0.03050 = 6.0·10⁻⁴ [m·mm]
a = 6.0·10⁻⁴ m·mm = 6.0·10⁻⁴ · 10⁻³ = 6.0·10⁻⁷ [] λ = 6.0·10⁻⁷2 · 0.75 = 6.0·10⁻⁷1.5 = 4.0·10⁻⁷ m
a = 6.20·10⁻⁴ m·mm  ⟹  λ = 6.20·10⁻⁷1.5 ≈ 4.13·10⁻⁷ m = 413 nm a = 6.79·10⁻⁴ m·mm  ⟹  λ = 6.79·10⁻⁷1.5 ≈ 4.53·10⁻⁷ m = 453 nm

x_min = ± m·λ·Lw
x_max = m·λ·Ld  ;  x_min = (m + 12)·λ·Ld
d = w  ⟹  λ·Ld = λ·Lw

h = 6.626·10−34 J·s  ;  c = 3·108 m/s  ;  e = 1.6·10−19 C
1 eV = 1.6·10−19 J  ;  h·c = 1242 eV·nm
E = h·cλ  ;  e·V = h·cλ − B  ;  λ0 = h·cB = 688.88 nm
500 nm
100 %
0.0 V
2.48 eV
1.8 eV
0.68 eV
0.68 V
0.0
1.20 V

E = h·cλ
h·cλ ≥ B λ ≤ h·cB
λ0 = h·cB = 688.88 nm

I = ΔQΔt = N·eΔt
Ek,max = h·cλ − B

Ek,max = h·cλ − B

e·V = Ek,max = h·cλ − B
h·cλ0 = B λ0 = h·cB = 688.88 nm

e·V = h·cλ − B V = h·ce·1λBe
V = − Be
B = h·cλ0 = 6.626·10−34 · 3·108688.88·10−9 = 2.886·10−19 J
B = 2.886·10−191.6·10−19 = 1.8 eV Be = − 2.886·10−191.6·10−19 = −1.8 V

1λ1 < 1λ0 < 1λ2 < 1λ3
V(λ) = h·ce·1λBe > V1
h·cλ1 < B
V(λ2) < V1 ⟹ Ek,max2) < e·V1
V(λ3) > V1 ⟹ Ek,max3) > e·V1

n = 1 : E1 = −10.4 eV    n = 2 : E2 = −5.5 eV    n = 3 : E3 = −3.7 eV
n = 4 : E4 = −1.6 eV    n = ∞ : E = 0 eV
h = 6.626·10−34 J·s  ;  c = 3·108 m/s  ;  1 eV = 1.6·10−19 J  ;  h·c = 1242 eV·nm
Eph = h·cλ  ;  λ = h·cΔE
254 nm
4.89 eV
7.8·10⁻¹⁹
n = 1
0

Eph = ΔE = Eסופית − E1
λ = h·cEph ⟹ λmax ⇔ Eph,min
ΔE = E2 − E1 = (−5.5) − (−10.4) = 4.9 eV ΔE = 4.9 · 1.6·10−19 = 7.84·10−19 J
λmax = h·cΔE = 6.626·10−34 · 3·1087.84·10−19 = 2.54·10−7 m λmax = 12424.9 ≈ 254 nm

N = n·(n − 1)2 ; n = 3 ⟹ N = 3·22 = 3
3 → 2 : ΔE = (−3.7) − (−5.5) = 1.8 eV 2 → 1 : ΔE = (−5.5) − (−10.4) = 4.9 eV 3 → 1 : ΔE = (−3.7) − (−10.4) = 6.7 eV
λ(3 → 2) = 12421.8 = 690 nm λ(2 → 1) = 12424.9 ≈ 254 nm λ(3 → 1) = 12426.7 ≈ 185 nm

1 → 2 : ΔE = 4.9 eV ⟹ λ = 12424.9 ≈ 254 nm 1 → 3 : ΔE = 6.7 eV ⟹ λ = 12426.7 ≈ 185 nm
1.8 eV ≠ En − E1

Eph = h·cλ = 6.626·10−34 · 3·10880·10−9 = 2.485·10−18 J Eph = 2.485·10−181.6·10−19 ≈ 15.5 eV
Eיינון = E − E1 = 0 − (−10.4) = 10.4 eV 15.5 > 10.4 ⟹ ✔
Ek,max = Eph − Eיינון = 15.5 − 10.4 = 5.1 eV Ek,max = 5.1 · 1.6·10−19 = 8.2·10−19 J

Eph = ΔE  ;  Ee ≥ ΔE
1 → 2 : ΔE = 4.9 eV ≤ 8 ✔ 1 → 3 : ΔE = 6.7 eV ≤ 8 ✔ 1 → 4 : ΔE = 8.8 eV > 8 ✘
6.7 < 8 < 8.8 ⟹ nmax = 3

t (days)  |  0  |  8  |  16  |  24  |  32
R (Bq)  |  900  |  450  |  225  |  112  |  56
R0 = 900 Bq  ;  1 Bq = 1 decay/s  ;  1 day = 86400 s
ln2 = 0.693  ;  R = λ·N  ;  λ = ln2
×40
8.0 days
450 Bq
59.9 days
0.0
900
900
0.00
1.003·10⁻⁶
8.97·10⁸
131
53
78

p–p  ·  p–n  ·  n–n
range ≈ 10⁻¹⁵ m
FCoulomb = k·q₁·q₂

¹³¹₅₃I → ¹³¹₅₄Xe* + ⁰₋₁e + ν̄ A : 131 = 131 + 0 Z : 53 = 54 + (−1)
¹³¹₅₄Xe*
¹³¹₅₄Xe* → ¹³¹₅₄Xe + γ A : 131 = 131    Z : 54 = 54
¹³¹₅₃I → ¹³¹₅₄Xe* + ⁰₋₁e (+ ν̄) → ¹³¹₅₄Xe + γ

R₀ = 900 Bq
R(T½) = R₀2 = 9002 = 450 Bq
T½ = 8 days
R(16) = 9004 = 225 Bq R(24) = 9008 = 112.5 Bq

R = λ·N   ⇒   N₀ = R₀λ
T½ = 8 · 24 · 3600 = 6.912·10⁵ s
λ = ln2 = 0.6936.912·10⁵ = 1.003·10⁻⁶ 1/s
N₀ = R₀λ = 9001.003·10⁻⁶ = 8.97·10⁸ ≈ 9·10⁸

R = R₀ · 2−n  ,  n = t
2n = R₀R = 9005 = 180
t = log₂180 = ln180ln2 = 5.1930.693 = 7.49
t = 7.49 · 8 = 59.9 ≈ 60 days t = ln2 · ln180 = 11.54 · 5.193 = 59.9 days

n → p + ⁰₋₁e + ν̄
N = N₀ · 2−n  ,  n = t
F = q·v·B·sinθ  ;  γ : q = 0 ⇒ F = 0
max EbA at A ≈ 56
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