פתרון בגרות · קרינה וחומר

✦ בנוי על ידי פהד גאנם ✦

A = 0.2 m  ·  λ = 0.4 m  ·  T = 0.8 s  ·  f = 1.25 Hz
d1 = 32 cm  ·  dr = 32 cm  ·  dt = 16 cm
0.00 s
0.00 s
0.000 m
0.00 m/s
0.000 m
0.00 m/s
0.50 m/s
y(x,t) = A·cos(k·x − ω·t) vy = −v · ∂y∂x r = v₂ − v₁v₂ + v₁ = − 13 τ = 2·v₂v₁ + v₂ = 23

ymax = +0.2 m  ;  ymin = −0.2 m
2A = 0.4 m  ;  A = 0.42 = 0.2 m

λ = 0.4 m
T = 1.0 − 0.2 = 0.8 s f = 1T = 10.8 = 1.25 Hz
v₁ = λT = 0.40.8 = 0.5 m/s
v₁ = λ·f = 0.4 · 1.25 = 0.5 m/s

vy = −v · ∂y∂x
x = 0.1 : ∂y∂x = −π  ;  vy = −0.5·(−π) = +1.57 m/s
x = 0.3 : ∂y∂x = +π  ;  vy = −0.5·(+π) = −1.57 m/s

v₁ = 0.5 m/s
tj = d₁v₁ = 0.320.5 = 0.64 s
Δt = 0.320.5 = 0.64 s
t₁ = 0.64 + 0.64 = 1.28 s t₁ = 0.640.5 = 1.28 s

v = √FTμ
μ₂μ₁ = v₁²v₂² = 2² = 4

Δt = 1.28 − 0.64 = 0.64 s
v₂ = ΔxΔt = 0.160.64 = 0.25 m/s
v₂v₁ = 1632 = 0.5  ;  v₂ = 0.5 · 0.5 = 0.25 m/s

d = 0.5 mm  ·  L = 2.5 m  ·  xV = 2.0 mm  ·  xR = 3.5 mm
N* = 7000 1/cm = 7·10⁵ 1/m  ·  n = 1
2.5 m
0.50 mm
0.0°
0.0°
0.0000
0.0000
0.00
0.00
0.00
0.00
0.00
0.00
0.0%
0.00
0.00
0.0%
0.00
0.00
0.0%
400 nm
0.0°
0.00
d·sin θ = n·λ tan θ = xL d = 1N* Δx = Ld · (λR − λV)

d·sin θ = n·λ  ;  n = 1
sin θ ≈ tan θ = xL ≤ 1.4·10⁻³
λ = d·xn·L
λV = 0.5·10⁻³ · 2·10⁻³1 · 2.5 = 4.00·10⁻⁷ m = 400 nm
λR = 0.5·10⁻³ · 3.5·10⁻³1 · 2.5 = 7.00·10⁻⁷ m = 700 nm
λRλV = 700400 = 1.75 = 3.52.0

Δ = 0 = n·λ  ;  n = 0  ;  ∀λ
sin θ = λd  ;  x = L·λd

x = L·λd
Δx = xR − xV = Ld · (λR − λV)
Δx = 2.50.5·10⁻³ · 3·10⁻⁷ = 1.5·10⁻³ m = 1.5 mm

d = 1N* = 17·10⁵ = 1.4286·10⁻⁶ m
sin θ = λd = λ·N* sin θV = 4·10⁻⁷ · 7·10⁵ = 0.28  ;  θV = 16.3° sin θR = 7·10⁻⁷ · 7·10⁵ = 0.49  ;  θR = 29.3°
tan θ = xL  ;  x = L·tan θ
cos θV = √1 − 0.28² = √0.9216 = 0.96 tan θV = 0.280.96 = 0.29167 cos θR = √1 − 0.49² = √0.7599 = 0.87172 tan θR = 0.490.87172 = 0.56211
xV = 2.5 · 0.29167 = 0.729 m xR = 2.5 · 0.56211 = 1.405 m Δx = 1.405 − 0.729 = 0.676 m
S = 2·xR = 2 · 1.405 = 2.81 m

λ1 = 410 nm  |  λ2 = 490 nm
Vstop(ג) = Vstop(ב) = 0.82 V  |  Vstop(א) = 0.33 V
h = 6.626·10⁻³⁴ J·s  ;  c = 3·10⁸ m/s  ;  e = 1.6·10⁻¹⁹ C
h·c = 1.9878·10⁻²⁵ J·m  ;  me = 9.1·10⁻³¹ kg
0.00 V
410 nm
100 %

h·f = W + Ek,max

e·Vstop = Ek,max = h·f − W
ג , ב : V = −0.82 V    א : V = −0.33 V
E = h·cλ  ;  0.82 V > 0.33 V

h·cλ = W + e·Vstop
λ = 410 nm = 4.10·10⁻⁷ m  ;  Vstop = 0.82 V
E = h·cλ = 1.9878·10⁻²⁵4.10·10⁻⁷ = 4.848·10⁻¹⁹ J = 3.03 eV
Ek,max = e·Vstop = 1.6·10⁻¹⁹ · 0.82 = 1.312·10⁻¹⁹ J = 0.82 eV
W = E − Ek,max = 4.848·10⁻¹⁹ − 1.312·10⁻¹⁹ = 3.536·10⁻¹⁹ J W ≈ 3.53·10⁻¹⁹ J ≈ 2.21 eV
E = 1.9878·10⁻²⁵4.90·10⁻⁷ = 4.057·10⁻¹⁹ J W = 4.057·10⁻¹⁹ − 1.6·10⁻¹⁹·0.33 = 4.057·10⁻¹⁹ − 0.528·10⁻¹⁹ = 3.529·10⁻¹⁹ J

Ek,max = h·f − W = 1.31·10⁻¹⁹ J = 0.82 eV vmax = √2·Ek,maxme ≈ 5.4·10⁵ m/s

I = n · e

150 nm ≤ λ ≤ 350 nm  |  : 150 nm ≤ λ ≤ 177 nm
: 214 nm   225 nm   248 nm   310 nm
h = 6.626·10⁻³⁴ J·s  ;  c = 3·10⁸ m/s  ;  e = 1.6·10⁻¹⁹ C
h·c = 1.9878·10⁻²⁵ J·m = 1242.375 eV·nm
150 nm

n = 1
Ephoton = Em − E1
Ek = h·cλ − (E∞ − E1)
E∞ − E1 = 1.9878·10⁻²⁵177·10⁻⁹ = 1.123051·10⁻¹⁸ J = 7.02 eV

310 nm : 1 → 2  ·  248 nm : 1 → 3  ·  225 nm : 1 → 4  ·  214 nm : 1 → 5  ·  177 nm : 1 → ∞
En = h·cλ E₂ = 1.9878·10⁻²⁵310·10⁻⁹ = 6.412258·10⁻¹⁹ J = 4.01 eV E₃ = 1.9878·10⁻²⁵248·10⁻⁹ = 8.015323·10⁻¹⁹ J = 5.01 eV E₄ = 1.9878·10⁻²⁵225·10⁻⁹ = 8.834667·10⁻¹⁹ J = 5.52 eV E₅ = 1.9878·10⁻²⁵214·10⁻⁹ = 9.288785·10⁻¹⁹ J = 5.81 eV E∞ = 1.9878·10⁻²⁵177·10⁻⁹ = 1.123051·10⁻¹⁸ J = 7.02 eV
En(3) = En(2) − 1.123051·10⁻¹⁸ J E₁ = −1.123051·10⁻¹⁸ J = −7.02 eV E₂ = 6.412258·10⁻¹⁹ − 1.123051·10⁻¹⁸ = −4.818252·10⁻¹⁹ J = −3.01 eV E₃ = 8.015323·10⁻¹⁹ − 1.123051·10⁻¹⁸ = −3.215187·10⁻¹⁹ J = −2.01 eV E₄ = 8.834667·10⁻¹⁹ − 1.123051·10⁻¹⁸ = −2.395843·10⁻¹⁹ J = −1.50 eV E₅ = 9.288785·10⁻¹⁹ − 1.123051·10⁻¹⁸ = −1.941725·10⁻¹⁹ J = −1.21 eV E∞ = 0
6.412 > 1.603 > 0.819 > 0.454   (10⁻¹⁹ J)

214 nm  ·  225 nm  ·  248 nm  ·  310 nm
ΔE = E₅ − E₂ = 9.288785·10⁻¹⁹ − 6.412258·10⁻¹⁹ = 2.876527·10⁻¹⁹ J = 1.798 eV λ = h·cΔE = 1.9878·10⁻²⁵2.876527·10⁻¹⁹ = 6.9104·10⁻⁷ m ≈ 691 nm
1λ = 1214 − 1310  ⇒  λ = 214 · 31096 = 6634096 = 691.0 nm
ΔE = 8.834667·10⁻¹⁹ − 6.412258·10⁻¹⁹ = 2.422409·10⁻¹⁹ J = 1.514 eV λ = 1.9878·10⁻²⁵2.422409·10⁻¹⁹ = 8.206·10⁻⁷ m = 820.6 nm
3 → 2 : ΔE = 1.603065·10⁻¹⁹ J  ⇒  λ = 1240 nm 5 → 3 : ΔE = 1.273462·10⁻¹⁹ J  ⇒  λ = 1561 nm 4 → 3 : ΔE = 8.19344·10⁻²⁰ J  ⇒  λ = 2426 nm 5 → 4 : ΔE = 4.54118·10⁻²⁰ J  ⇒  λ = 4377 nm

Ephoton = 1.9878·10⁻²⁵150·10⁻⁹ = 1.325200·10⁻¹⁸ J = 8.28 eV
E∞ − E₁ = 1.123051·10⁻¹⁸ J = 7.02 eV
Ephoton = (E∞ − E₁) + Ek Ek = 1.325200·10⁻¹⁸ − 1.123051·10⁻¹⁸ = 2.021490·10⁻¹⁹ J Ek = 2.021490·10⁻¹⁹1.6·10⁻¹⁹ = 1.26 eV

3T½ = 24 days  |  N(3T½) = 0.4·108 = 4·107  |  T½ = 8 days
N(t) = N0·e−λ·t     λ = ln 2T½     A = λ·N     NN0 = AA0 = e−λ·t
0.00 days
8.0 days
1.0028·10⁻⁶ 1/s
3.20·10⁸
26.58 days
3.20·10⁸
100.00%
320.9 Bq
100.00%
0.00 days
0.00·10⁰ s
0.000
400 / 400
8.0 days
6.912·10⁵ s
1.0028·10⁻⁶
0.086643
26.58 days
2.296·10⁶ s
A0 = λ·N0 = 320.9 Bq NN0 = AA0 = e−λ·t

N(3T½) = 0.4·108 = 4·107
N(3T½) = N023 = N08
N0 = 8 · N(3T½) = 8 · 4·107 = 3.2·108
N = N0 · 2−n  ,  n = tT½ = 248 = 3
3.2·108 → 1.6·108 → 0.8·108 → 0.4·108

T½ = 24 days3 = 8 days
12 = e−λ·T½ ⟹ λ·T½ = ln 2 ⟹ λ = ln 2T½
T½ = 8 · 24 · 60 · 60 = 8 · 86 400 = 691 200 s
λ = ln 2T½ = 0.693147691 200 s = 1.0028·10−6 1/s
1.0028171·10−6 · 691 200 = 0.693147 ✓ 0.0866434 1/day86 400 s/day = 1.0028·10−6 1/s ✓

A = λ·N  ,  A0 = λ·N0 ⟹ AA0 = λ·Nλ·N0 = NN0
AA0 = 110 ⟺ e−λ·t = 110
−λ·t = ln110 = −ln 10 ⟹ t = ln 10λ
t = 2.3025851.0028171·10−6 1/s = 2.2961·106 s
t = 2.2961·106 s86 400 s/day = 26.58 days
t = T½ · log210 = 8 · 3.3219 = 26.58 days
24 days < 26.58 days < 32 days  |  6.25% < 10% < 12.5% ✓

N = N0·e−λ·t ⟹ NN0 = e−λ·t
AA0 = λ·Nλ·N0 = e−λ·t
2·A2·A0 = AA0
t = ln 10λ = T½·log210 = 26.58 days

n = 365 days8 days = 45.6 ⟹ NN0 = 2−45.6 ≈ 2·10−14
T½ = ln 2λ ⟹ λ ↑ ⟹ T½ ↓
λ(I-131) = 1.0028·10−6  ;  λ(C-14) = 3.8332·10−12 1/s
I-131 : t0.1 = 26.58 days  |  C-14 : t0.1 = 19 035 years
| פהד גאנם
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