פתרון בגרות · קרינה וחומר

✦ בנוי על ידי פהד גאנם ✦


u  m  |   0.13  |  0.18  |  0.25  |  0.33  |  0.60
v  m  |   0.44  |  0.22  |  0.16  |  0.14  |  0.12
1u   ?   |   ?   |   ?   |   ?   |   ?   |   ?
1v   ?   |   ?   |   ?   |   ?   |   ?   |   ?
1u + 1v = 1f
0.250 m
0.160 m
1.00 R
18
0.250
0.160
4.00
6.25
10.25
-0.64
100%
100%

1u + 1v = 1f
1v = 1f1u 1v = −1u + 1f
x = 1u  ;  y = 1v  ;  y = m·x + b m = −1  ;  b = 1f
v = u·fu − f

1u = 10.13 = 7.69 1/m 1v = 10.44 = 2.27 1/m
10.13 = 7.69  ;  10.18 = 5.56  ;  10.25 = 4.00  ;  10.33 = 3.03  ;  10.60 = 1.67  1/m
10.44 = 2.27  ;  10.22 = 4.55  ;  10.16 = 6.25  ;  10.14 = 7.14  ;  10.12 = 8.33  1/m

(7.69 , 2.27)   (5.56 , 4.55)   (4.00 , 6.25)   (3.03 , 7.14)   (1.67 , 8.33)
7.69 + 2.27 = 9.97  ;  5.56 + 4.55 = 10.10  ;  4.00 + 6.25 = 10.25  ;  3.03 + 7.14 = 10.17  ;  1.67 + 8.33 = 10.00  1/m 9.97 + 10.10 + 10.25 + 10.17 + 10.005 = 10.10 1/m

1u = 0   ⇒   1v = 1f f = 1b
1v = 0   ⇒   1u = 1f
m = −1.015 ≈ −1  ;  b = 1f = 10.16 1/m f = 110.16 = 0.098 m
1f = 10.10 1/m   ⇒   f = 110.10 = 0.099 m

1v = 1f1u
II0 = π·r2π·R2 = r2R2

λ1 = 600 nm = 6·10−7 m  ;  L = 1.5 m
Δx = λ·Ld  ;  W = 2·λ·Lw
: Δr = n·λ  ;  : Δr = (2n+1)·λ2
600 nm
60 µm
1.50 m
60 µm
1.20
3.75 cm
1.50
1.25
3.75
2.50
3.00
2.50 λ
3.00
2.00

Δr = 18 · λ12 = 182·λ1 = 9λ1
Δr = n·λ  ;  n = 0, 1, 2, ... Δr = (2n+1)·λ2
n = Δrλ1 = 1λ1 = 9

8 − 0 = 8  ;  12 cm = 8·Δx
Δx = 12 cm8 = 1.5 cm
Δx = 1.5 cm = 1.5·10−2 m

Δx = λ·Ld
Δx2Δx1 = λ2λ1
λ2 = λ11.2 = 600 nm1.2 = 500 nm
Δx2 = 1.5 cm1.2 = 1.25 cm

n = yΔx  ;  Δr = n·λ
n1 = 3.75 cm1.5 cm = 2.5 Δr1 = 2.5·λ1 = 5·λ12
n2 = 3.75 cm1.25 cm = 3 Δr2 = 3·λ2

w·sinθ = λ1  ⇒  y1 = λ1·Lw W = 2·λ1·Lw
Δx = λ1·Ld
λ1·Lw = 2·λ1·Ld 1w = 1d
w = d

w = d
Δx = λ1·Ld  ⇒  d = λ1·LΔx
d = 6·10−7 · 1.51.5·10−2 = 9·10−71.5·10−2 = 6·10−5 m
w = d = 6·10−5 m = 0.06 mm W = 2·6·10−7 · 1.56·10−5 = 0.03 m = 3 cm = 2 · 1.5 cm

c = 3·108 m/s  ·  h = 6.626·10−34 J·s  ·  e = 1.6·10−19 C
1 eV = 1.6·10−19 J  ·  h·c = 1242 eV·nm
V(f = 0) = −1.25 V  ·  V2(I = 0) = −0.85 V  ·  f0 < f2 < f1
1.25 eV
1.25 eV
2.0·10⁻¹⁹ J
3.02·10¹⁴ Hz
994 nm
4.14·10⁻¹⁵ V·s
−1.25 V

f ≥ f0 f < f0
h·f0 = B Ek,max = h·f0 − B = 0
V(f0) = 0

Ek,max = h·f − B e·V = h·f − B V = he · f − Be
m = he n = V(f = 0) = − Be
Be = −1.25 V B = e · 1.25 V = 1.25 eV
B = 1.25 · 1.6·10−19 = 2.0·10−19 J
|n| = Be = 1.25 V

λ = cf h·f0 = B
h · cλmax = B λmax = h·cB
λmax = 6.626·10−34 · 3·1082.0·10−19 = 1.9878·10−252.0·10−19 = 9.94·10−7 m
λmax = 1242 eV·nm1.25 eV ≈ 994 nm

Vs = 0.85 V
Ek,max = e·Vs = 0.85 eV
h·f1 = B + Ek,max = 1.25 + 0.85 = 2.10 eV h·f1 = 2.10 · 1.6·10−19 = 3.36·10−19 J
f1 = 3.36·10−196.626·10−34 = 5.07·1014 Hz
λ1 = 3·1085.07·1014 ≈ 5.92·10−7 m ≈ 592 nm

Ek,max = e·Vs Vs = 0.85 V
Ek,max = 1.6·10−19 · 0.85 = 1.36·10−19 J
Ek,max = h·f1 − B = 3.36·10−19 − 2.0·10−19 = 1.36·10−19 J

Ek,max = h·f − B f2 < f1 ⟸ h·f2 − B < h·f1 − B
Vs = h·f1 − Be = 0.85 V
f2 > f0 ⟸ h·f2 > B
Itot(V = 0) = I1 + I2 > I1

c = 3·108 m/s  ·  h = 6.626·10−34 J·s  ·  1 eV = 1.6·10−19 J
h·c = 1242 eV·nm  ·  λion = 200 nm  ·  Ek = 1.06 eV
λabs = 285 , 330 , 589 nm  ·  400 nm < λ < 700 nm
200 nm
1.0×
6.21 eV
9.94·10⁻¹⁹ J
5.15 eV
1.06 eV
241 nm

Na + E → Na+ + e
Eion = 0 − E1 = |E1|
Eph = Eion + Ek

Eph = h·cλ = 1242 eV·nm200 nm = 6.21 eV
Eph = 6.21 · 1.6·10−19 = 9.94·10−19 J Eph = 6.626·10−34 · 3·1082·10−7 = 9.94·10−19 J
Eion = Eph − Ek = 6.21 − 1.06 = 5.15 eV
Eion = 5.15 · 1.6·10−19 = 8.24·10−19 J

ΔE = En − E1 = h·cλ
ΔE(589) = 1242589 = 2.11 eV ΔE(330) = 1242330 = 3.76 eV ΔE(285) = 1242285 = 4.36 eV
E = 0 E1 = −Eion = −5.15 eV
E2 = −5.15 + 2.11 = −3.04 eV E3 = −5.15 + 3.76 = −1.39 eV E4 = −5.15 + 4.36 = −0.79 eV

ΔE(2→1) = 2.11 eV λ = 12422.11 = 589 nm

Eph = Ehigh − Elow
N = 4 · 32 = 6
4→3 , 4→2 , 4→1 , 3→2 , 3→1 , 2→1

λ = h·cΔE = 1242ΔE
2→1 : ΔE = −3.04 − (−5.15) = 2.11 eV ⟸ λ = 589 nm 3→1 : ΔE = −1.39 − (−5.15) = 3.76 eV ⟸ λ = 330 nm 4→1 : ΔE = −0.79 − (−5.15) = 4.36 eV ⟸ λ = 285 nm
4→2 : ΔE = −0.79 − (−3.04) = 2.25 eV ⟸ λ = 12422.25 = 552 nm 3→2 : ΔE = 1.65 eV ⟸ λ ≈ 750 nm 4→3 : ΔE = 0.59 eV ⟸ λ ≈ 2090 nm
589 nm , 552 nm

²²⁶₈₈Ra  ⟶  ²¹⁴₈₄Po   ;   ²²²₈₆Rn  ⟶  Y + ⁴₂He
A₀A = 8   ;   t = 11.475 d   ;   1 d = 86400 s
EB(Ra) = 1732.62 MeV   ;   EB(Po) = 1666.02 MeV
ln2 = 0.693   ;   λ = ln2
²²⁶₈₈Ra
1.00 ×
226
88
138
0
0
8
11.475 d
3.825 d
0.181
3.00
8.00
226 , 88 , 138
214 , 84 , 130
0.118
²¹⁴₈₄Po

α : ΔA = −4 , ΔZ = −2   (⁴₂He) β⁻ : ΔA = 0 , ΔZ = +1   (⁰₋₁e)
ΔA = 226 − 214 = 12 Nα = ΔA4 = 124 = 3
Z (3α) = 88 − 2·3 = 82 Nβ = 84 − 82 = 2
²²⁶₈₈Ra ⟶ ²¹⁴₈₄Po + 3·⁴₂He + 2·⁰₋₁e A : 226 = 214 + 12 ✓ Z : 88 = 84 + 6 − 2 ✓
²²⁶₈₈Ra ⟶(α) ²²²₈₆Rn ⟶(α) ²¹⁸₈₄Po ⟶(α) ²¹⁴₈₂Pb ⟶(β⁻) ²¹⁴₈₃Bi ⟶(β⁻) ²¹⁴₈₄Po

²²²₈₆Rn ⟶ Y + ⁴₂He
222 = AY + 4 AY = 222 − 4 = 218
86 = ZY + 2 ZY = 86 − 2 = 84
²²²₈₆Rn ⟶ ²¹⁸₈₄Po + ⁴₂He N(Y) = 218 − 84 = 134

A = λ·N A₀A = N₀N = 8
N = N₀2ⁿ N₀N = 2ⁿ = 8 = 2³  ⟹  n = 3
t = n·T½ T½ = tn = 11.4753 = 3.825 d
T½ = 3.825 · 24 · 3600 = 3.3048·10⁵ s ≈ 3.3·10⁵ s
N = N₀·e−λt  ⟹  λ·T½ = ln2 λ = ln2 = 0.6933.3048·10⁵ = 2.10·10⁻⁶ s⁻¹ λ = 0.6933.825 = 0.181 d⁻¹

EBA
(EBA)Ra = 1732.62226 = 7.667 MeV
(EBA)Po = 1666.02214 = 7.785 MeV
7.785 > 7.667 Δ(EBA) = 7.785 − 7.667 = 0.118 MeV
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