פתרון בגרות · קרינה וחומר

✦ בנוי על ידי פהד גאנם ✦

S1A = 29 cm  ;  S2A = 30 cm  ;  n = 1
v = 3 cm/s  ;  dא = dב  ;  A1 = A2 = A
|r2 − r1| = n·λ  ;  sinθ = n·λd  ;  v = λ·f
1.00 cm
4.0 cm
3.0 cm/s
1.00
3.00
3.00
0.333
0.00
0.00
0.00
0.00
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0.250
14.5
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4
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λ1 < λ2
v1 = v2 = v
v = λ·f f = vλ f1 = vλ1  ;  f2 = vλ2 f1 > f2
sinθ = n·λd

|r2 − r1| = n·λ n = 0, 1, 2, 3, …
Δr = |S2A − S1A| Δr = |30 − 29| = 1 cm
λ = Δrn = 11 = 1 cm

Δr = n·λ Δr = (n + 12)·λ
Δr(B) = 1·λ = 1 cm AB = A + A = 2A
Δr(C) = λ2 = 0.5 cm AC = A − A = 0

v = λ·f f = vλ
f = 3 cm/s1 cm = 3 Hz
T = 1f = 13 s

sinθ1 = λd
λ = vf

L = 1.5 m  ;  λred − λblue = 200 nm  ;  Δxblue = 0.8 cm
NA = 5  ;  NB = 7  ;  c = 3·108 m/s
Δx = λ·Ld  ;  xm = m·λ·Ld  ;  d·sinθ = m·λ
500 nm
0.09375 mm
1.5 m
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Δx = λ·Ld
Δx = Ld · λ
ΔxAΔxB = 75 = 1.4

5·Δxred = 7·Δxblue λredλblue = 75 = 1.4
λred − λblue = 200 nm λred = 1.4·λblue
1.4·λblue − λblue = 0.4·λblue = 200 nm λblue = 200 nm0.4 = 500 nm
λred = 1.4 · 500 = 700 nm 700 − 500 = 200 nm 700500 = 1.4

Δx = λ·Ld d = λ·LΔx
λblue = 500 nm = 5·10−7 m Δxblue = 0.8 cm = 8·10−3 m L = 1.5 m
d = 5·10−7 · 1.58·10−3 = 7.5·10−78·10−3 = 9.375·10−5 m
Δxred = 7·10−7 · 1.59.375·10−5 = 1.12·10−2 m = 1.12 cm

Δx = λ·Ld
d′ = 2d Δx′ = λ·L2d = Δx2

f1 = f2 λ = cf
Δφ = φ1 − φ2 = const

λ (nm)  |  520  |  500  |  480  |  440  |  400
V (V)   |  0.2  |  0.3  |  0.4  |  0.6  |  0.9
c = 3·108 m/s  ;  e = 1.6·10−19 C  ;  1 eV = 1.6·10−19 J
520 nm
60 %
0.00 V
520 nm
5.77·10¹⁴ Hz
2.39 eV
0.00 V
12.0 nA
0.20 V
1.21·10⁻⁶ V·m
6.4·10⁻³⁴ J·s
3 %
−2.1 V
2.1 eV
3.4·10⁻¹⁹ J
570 nm
5.25·10¹⁴ Hz
Eph = B + Ek,max
Eph = h·f = h·cλ  ;  Ek,max = e·V
V = h·ce · 1λ − Be
a = h·ce  ⇒  h = a·ec
f0 = Bh = cλ0

Eph = B + Ek,max
Eph = h·f  ;  c = λ·f  ⇒  f = cλ Eph = h·cλ
Ek,max = e·V
e·V = h·cλ − B V = h·ce·λ − Be = h·ce · 1λ − Be

1λ = 1520·10−9 = 1.92·106 m⁻¹ 1λ = 1400·10−9 = 2.50·106 m⁻¹
(1.92·106 , 0.2)   (2.00·106 , 0.3)   (2.08·106 , 0.4)   (2.27·106 , 0.6)   (2.50·106 , 0.9)
a = V2 − V1(1λ)2 − (1λ)1

a = h·ce
a = 0.9 − 0.22.50·106 − 1.92·106 = 0.70.58·106 ≈ 1.21·10−6 V·m
h = a·ec = 1.21·10−6 · 1.6·10−193·108 = 1.93·10−253·108 ≈ 6.4·10−34 J·s
6.63·10−34 − 6.44·10−346.63·10−34 ≈ 0.03

V0 = h·ce · 0 − Be = − Be
V0 = V1 − a · (1λ)1 = 0.2 − 1.21·10−6 · 1.92·106 V0 = 0.2 − 2.32 ≈ −2.1 V
B = e · |V0| = 1.6·10−19 · 2.1 ≈ 3.4·10−19 J = 2.1 eV

h·f0 = B + 0  ⇒  f0 = Bh
f0 = 2.1 · 1.6·10−196.4·10−34 = 3.36·10−196.4·10−34 ≈ 5.25·1014 Hz
0 = a · 1λ0 − Be  ⇒  1λ0 = 2.11.21·10−6 ≈ 1.74·106 m⁻¹ λ0 = 11.74·106 ≈ 5.75·10−7 m = 575 nm f0 = cλ0 = 3·1085.75·10−7 ≈ 5.2·1014 Hz

n  =  ∞  |  4  |  3  |  2  |  1
E (eV)  |  0  |  −1.6  |  −3.7  |  −5.5  |  −10.4
h = 6.626·10−34 J·s  ;  c = 3·108 m/s  ;  1 eV = 1.6·10−19 J
5.5 eV
5.5 eV
4.9 eV
0.6 eV
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6
2
4→3  |  2.1 eV  |  3.36·10−19 J  |  592 nm
3→2  |  1.8 eV  |  2.88·10−19 J  |  690 nm
4→2  |  3.9 eV  |  6.24·10−19 J  |  319 nm
2→1  |  4.9 eV  |  7.84·10−19 J  |  254 nm
3→1  |  6.7 eV  |  1.07·10−18 J  |  185 nm
4→1  |  8.8 eV  |  1.41·10−18 J  |  141 nm
Eph = Eסופית − Eהתחלתית
E = h·f = h·cλ  ⇒  λ = h·cE
h·c = 1.9878·10−25 J·m = 1.24·10−6 eV·m
N = n·(n−1)2 = 4·32 = 6

Eph = Efinal − Einitial
ΔE = E2 − E1 = (−5.5) − (−10.4) = 4.9 eV
Eph = 5.5 eV  ≠  ΔE = 4.9 eV
E3 − E1 = (−3.7) − (−10.4) = 6.7 eV E4 − E1 = (−1.6) − (−10.4) = 8.8 eV

Emin = E∞ − E1
Emin = 0 − (−10.4) = 10.4 eV
Emin = 10.4 · 1.6·10−19 = 1.66·10−18 J

Eph = Einitial − Efinal
N = 3 + 2 + 1 = 6
N = n·(n − 1)2 = 4·32 = 6
4→3 : (−1.6) − (−3.7) = 2.1 eV 4→2 : (−1.6) − (−5.5) = 3.9 eV 4→1 : (−1.6) − (−10.4) = 8.8 eV 3→2 : (−3.7) − (−5.5) = 1.8 eV 3→1 : (−3.7) − (−10.4) = 6.7 eV 2→1 : (−5.5) − (−10.4) = 4.9 eV

E = h·f  ;  f = cλ  ⇒  E = h·cλ  ;  λ = h·cE
h·c = 6.626·10−34 · 3·108 = 1.988·10−25 J·m E(700 nm) = 1.988·10−257·10−7 = 2.84·10−19 J = 1.77 eV E(400 nm) = 1.988·10−254·10−7 = 4.97·10−19 J = 3.1 eV
1.77 < 2.1 < 3.1 ✔    1.77 < 1.8 < 3.1 ✔ 3.9 , 4.9 , 6.7 , 8.8 > 3.1 ✘
λ43 = 1.988·10−253.36·10−19 = 5.92·10−7 m = 592 nm λ32 = 1.988·10−252.88·10−19 = 6.90·10−7 m = 690 nm

Eph = Efinal − E1
1→2 : (−5.5) − (−10.4) = 4.9 eV 1→3 : (−3.7) − (−10.4) = 6.7 eV 1→4 : (−1.6) − (−10.4) = 8.8 eV
Nemission = 6  ;  Nabsorption = 3

T½ = 88 y = 2.775·10⁹ s
N₀ = 2·10²⁵
t = 10 y = 3.154·10⁸ s
1 y = 365 · 24 · 3600 = 3.1536·10⁷ s
Z(Np) = 93 · Z(Pu) = 94 · A = 238
ln2 = 0.693
88
10
2·10²⁵
1.0×
2.50·10⁻¹⁰ s⁻¹
1.85·10²⁵
4.62·10¹⁵ Bq
1.5·10²⁴
92.4%
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A: 238 → 238 Z: 93 → 94
¹₀n → ¹₁p + ⁰₋₁e + ν̄
²³⁸₉₃Np → ²³⁸₉₄Pu + ⁰₋₁e + ν̄ A: 238 = 238 + 0 ✔ Z: 93 = 94 + (−1) ✔

T½ = 88 · 365 · 24 · 3600 = 88 · 3.1536·10⁷ s T½ = 2.775·10⁹ s
λ = ln2T½ = 0.6932.775·10⁹ = 2.50·10⁻¹⁰ 1/s
R₀ = λ · N₀ = 2.50·10⁻¹⁰ · 2·10²⁵ R₀ = 5.0·10¹⁵ Bq

t = 10 · 365 · 24 · 3600 = 3.154·10⁸ s λ · t = 2.50·10⁻¹⁰ · 3.154·10⁸ = 0.0788
N = N₀ · e−λ·t = 2·10²⁵ · e−0.0788 N = 2·10²⁵ · 0.924 = 1.85·10²⁵
N = N₀ · 2−tT½ = 2·10²⁵ · 2−1088 N = 2·10²⁵ · 0.924 = 1.85·10²⁵
ΔN = N₀ − N = 2·10²⁵ − 1.85·10²⁵ ≈ 1.5·10²⁴

N(T½) = N₀2 · N(2T½) = N₀4 · N(3T½) = N₀8 · N(4T½) = N₀16
1: 0.75 → 0.67 → 0.50 ✘ 2: 0.33 → 0.50 → 0.67 ✘ 4: 0.25 → 0.50 → 0.50 ✘
3: 0.50 → 0.50 → 0.50 ✔

α > β > γ α < β < γ
α = ⁴₂He · q = +2e
β = ⁰₋₁e · q = −1e
γ = photon · q = 0 · m₀ = 0
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