פתרון בגרות · קרינה וחומר

✦ בנוי על ידי פהד גאנם ✦

v_pulse = 2 cm/s ; = 4 cm ; = ±2 cm
: x ∈ [2 , 6] ; : x ∈ [10 , 14]
M : x = 6 cm ; N : x = 10 cm ; MN = 4 cm ; = 1 cm
L = 0.9 m ; v_string = 27 m/s ; B , C
f_n = n·v2·L = n · 15 Hz ; 0 ≤ f ≤ 70 Hz

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2 – 6 cm

10 – 14 cm

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: y(x , t) = y₁(x , t) + y₂(x , t) ; Δx = v · t

: L = n·λ2 ; λ_n = 2·Ln ; f_n = n·v2·L = n · 15 Hz

T = 1f ; : t = T2 = 12·f

y(x , t) = y₁(x , t) + y₂(x , t) Δx = v · t

Δx = v · t = 2 · 1 = 2 cm x_up(t) = [ 2 + 2t , 6 + 2t ] x_dn(t) = [ 10 − 2t , 14 − 2t ]

x_up(1) = [ 2 + 2 , 6 + 2 ] = [ 4 , 8 ] x_dn(1) = [ 10 − 2 , 14 − 2 ] = [ 8 , 12 ]

x_up(2) = [ 2 + 4 , 6 + 4 ] = [ 6 , 10 ] x_dn(2) = [ 10 − 4 , 14 − 4 ] = [ 6 , 10 ] y(x) = (+2) + (−2) = 0

x_up(3) = [ 2 + 6 , 6 + 6 ] = [ 8 , 12 ] x_dn(3) = [ 10 − 6 , 14 − 6 ] = [ 4 , 8 ]

y(x , t) = y₁(x , t) + y₂(x , t) = 0

t_in = 8 − 62 = 1 s ; t_out = 8 − 22 = 3 s t_in = 10 − 82 = 1 s ; t_out = 14 − 82 = 3 s

y(8 , t) = (+2) + (−2) = 0

= MN2 = 42 = 2 cm

v_y(x) ∝ − y(x)

y(x , t) = A · sin(k·x) · cos(ω·t)

y(x_node , t) = 0 ; v_y(x_node , t) = 0

: |v_y| y = 0 ; v_y = 0 |y| : |v_y| |y| ; v_y = 0 y = 0

L = n·λ2 , n = 1 , 2 , 3 , …

λ_n = 2·Ln = 2 · 0.9n = 1.8n m

v = λ · f ; f_n = vλ_n = n·v2·L

f_n = n · 272 · 0.9 = n · 271.8 = n · 15 Hz f₁ = 15 Hz ; f₂ = 30 Hz ; f₃ = 45 Hz ; f₄ = 60 Hz f₅ = 75 Hz > 70 Hz

n = ff₁ = 6015 = 4

λ = 2·Ln = 2 · 0.94 = 1.84 = 0.45 m λ = vf = 2760 = 0.45 m

= n = 4 ; λ2 = 0.452 = 0.225 m

= n + 1 = 4 + 1 = 5 x_node = 0 , 0.225 , 0.45 , 0.675 , 0.9 m

T = 1f = 115 ≈ 0.0667 s

t = T2

t = T2 = 1152 = 12 · 15 = 130 ≈ 0.033 s

d = 0.4 mm ; L = 3.4 m ; w = 0.2 mm ; x_min,1 = 2.55 mm
x_m = (m + ½)·λ·Ld x_n = n·λ·Ld Δy = λ·Ld Δx = 2·λ·Lw

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Δr = d·xL ; Δr = (m + ½)·λ₁ x_m = (m + ½)·λ₁·Ld

x_min,1 = λ₁·L2·d ⇐ λ₁ = 2·x_min,1·dL

λ₁ = 2 · 2.55·10⁻³ · 0.4·10⁻³3.4 = 2.04·10⁻⁶3.4 = 6.0·10⁻⁷ m λ₁ = 600 nm

Δy = λ₁·Ld = 6.0·10⁻⁷ · 3.40.4·10⁻³ = 5.1·10⁻³ m Δy2 = 2.55·10⁻³ m ✓

x_min,3 = (2 + ½)·λ₂·Ld = 2.5·λ₂·Ld

Δr = n·λ₁ ; x_max,2 = 2·λ₁·Ld

2.5·λ₂·Ld = 2·λ₁·Ld ⇐ 2.5·λ₂ = 2·λ₁ λ₂ = 2·λ₁2.5

λ₂ = 2 · 6002.5 = 12002.5 = 480 nm = 4.8·10⁻⁷ m

x = 2 · 6.0·10⁻⁷ · 3.40.4·10⁻³ = 1.02·10⁻² m x = 2.5 · 4.8·10⁻⁷ · 3.40.4·10⁻³ = 1.02·10⁻² m ✓

sinθ = n·λ₁w , n = 1, 2, 3 … sinθ₁ = λ₁w

tanθ₁ = xL ⇐ x = L·sinθ₁ = λ₁·Lw

Δx = 2·x = 2·λ₁·Lw

Δx = 2 · 6.0·10⁻⁷ · 3.40.2·10⁻³ = 4.08·10⁻⁶0.2·10⁻³ = 2.04·10⁻² m Δx = 20.4 mm

20.45.1 = 4

S₁ : x₁ + r₁ ; S₂ : x₂ + r₂ Δ_total = (x₂ + r₂) − (x₁ + r₁)

x₂ > x₁ ⇐ x₂ − x₁ > 0

x₁ + r₁ = x₂ + r₂ r₁ − r₂ = x₂ − x₁ > 0 ⇐ r₁ > r₂

O : r₁ = r₂ ⇐ Δ_total(O) = x₂ − x₁ ≠ 0

Δ_total = (x₂ + r₂) − (x₁ + r₁) r₁ = r₂ ⇐ Δ_total = x₂ − x₁

Δ_total = (m + ½)·λ₁ ; m = 1 x₂ − x₁ = (1 + ½)·λ₁ = 1.5·λ₁

x₂ − x₁ = 1.5 · 6.0·10⁻⁷ = 9.0·10⁻⁷ m = 900 nm

x₂ − x₁ = 9.0·10⁻⁷ m > 0 ✓

λ (nm) | 350 | 400 | 450 | 496 | 534
V (V) | 1.45 | 1.10 | 0.70 | 0.40 | 0.20
c = 3·10⁸ m/s ; e = 1.6·10⁻¹⁹ C ; h·c = 1242 eV·nm

0.70 eV

0.00 eV

450 nm

0.00 V

6.67·10¹⁴ Hz

2.76 eV

0.66 V

10.00 µA

4.30·10⁻¹⁵ V·s

6.88·10⁻³⁴ J·s

3.8 %

−2.19 V

2.19 eV

5.09·10¹⁴ Hz

589 nm

c = λ·f f = cλ

1 nm = 10⁻⁹ m λ = 350 nm = 350·10⁻⁹ m = 3.5·10⁻⁷ m

f = 3·10⁸350·10⁻⁹ = 8.57·10¹⁴ Hz f = 3·10⁸400·10⁻⁹ = 7.50·10¹⁴ Hz f = 3·10⁸450·10⁻⁹ = 6.67·10¹⁴ Hz f = 3·10⁸496·10⁻⁹ = 6.05·10¹⁴ Hz f = 3·10⁸534·10⁻⁹ = 5.62·10¹⁴ Hz

λ (nm): 350 | 400 | 450 | 496 | 534 f (10¹⁴ Hz): 8.57 | 7.50 | 6.67 | 6.05 | 5.62 V (V): 1.45 | 1.10 | 0.70 | 0.40 | 0.20

h·f = B + E_k,max E_k,max = h·f − B

E_k,max = e·V

e·V = h·f − B V = h·f − Be V = he·f − Be

y = m·x + n m = he ; n = −Be

f: 5·10¹⁴ Hz … 9·10¹⁴ Hz V: 0 V … 1.6 V

(8.57·10¹⁴ , 1.45) ; (7.50·10¹⁴ , 1.10) ; (6.67·10¹⁴ , 0.70) (6.05·10¹⁴ , 0.40) ; (5.62·10¹⁴ , 0.20)

V = m·f + n

m = ΔVΔf = V₂ − V₁f₂ − f₁

m = he ; h = m·e

m = 1.45 − 0.208.57·10¹⁴ − 5.62·10¹⁴ m = 1.252.95·10¹⁴ = 4.23·10⁻¹⁵ V·s

h = m·e = 4.23·10⁻¹⁵ · 1.6·10⁻¹⁹ h = 6.77·10⁻³⁴ J·s ≈ 6.8·10⁻³⁴ J·s

h_מקובל = 6.63·10⁻³⁴ J·s 6.8·10⁻³⁴ − 6.63·10⁻³⁴6.63·10⁻³⁴ ≈ 0.03

n = −Be = −2.19 V B = 2.19 · 1.6·10⁻¹⁹ = 3.5·10⁻¹⁹ J = 2.19 eV

B = h·cλ − e·V B = 1242350 − 1.45 = 3.55 − 1.45 = 2.10 eV

f₀ = Bh = 2.194.30·10⁻¹⁵ = 5.09·10¹⁴ Hz λ₀ = cf₀ = 3·10⁸5.09·10¹⁴ = 5.89·10⁻⁷ m = 589 nm

B(Cs) ≈ 2.1 eV ≈ 3.4·10⁻¹⁹ J

E_k,max = h·f − B = h·cλ − B

E_k,max = e·V = 1.6·10⁻¹⁹ · 0.7 = 1.12·10⁻¹⁹ J = 0.7 eV 1242450 − 2.1 = 2.76 − 2.1 = 0.66 eV ≈ 0.7 eV

E_הגעה = E_k,max − e·V = 0.7 − 0.5 = 0.2 eV = 3.2·10⁻²⁰ J

E_k,max(0.7 V) = E_k,max(0.5 V) = 0.7 eV = 1.12·10⁻¹⁹ J

: 100 nm – 119.2 nm
: 140.3 nm | 185.9 nm | 255.15 nm
h·c = 1242 eV·nm ; m = 9.109·10⁻³¹ kg ; 1 eV = 1.6·10⁻¹⁹ J

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10.42 eV

255.15 nm

4.87 eV

7.79·10⁻¹⁹ J

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0.00·10⁵ m/s

E_פוטון = h·cλ = E_m − E_n

λ_בליעה = h·cE_m − E_n

h·cλ = E_ion + E_k

E_ion = h·cλ_max

E_ion = 6.626·10⁻³⁴ · 3·10⁸119.2·10⁻⁹ = 1.67·10⁻¹⁸ J

E_ion = 1.67·10⁻¹⁸1.6·10⁻¹⁹ = 10.42 eV E_ion = 1242119.2 = 10.42 eV

ΔE = h·cλ = E_n − E₁

ΔE₂ = 1242255.15 = 4.87 eV ΔE₃ = 1242185.9 = 6.68 eV ΔE₄ = 1242140.3 = 8.85 eV

E_∞ = 0 eV ; E₁ = −E_ion = −10.42 eV

E₂ = −10.42 + 4.87 = −5.55 eV E₃ = −10.42 + 6.68 = −3.74 eV E₄ = −10.42 + 8.85 = −1.57 eV

E_ph = h·cλ_min

E_ph = 1242100 = 12.42 eV

E_k,max = E_ph − E_ion = 12.42 − 10.42 = 2.0 eV E_k,max = 2.0 · 1.6·10⁻¹⁹ = 3.2·10⁻¹⁹ J

E_k = m·v²2 v = √2·E_km v = √2 · 3.2·10⁻¹⁹9.109·10⁻³¹ = 8.4·10⁵ m/s

N = C(4,2) = 4·32 = 6

E(700 nm) = 1242700 = 1.77 eV E(400 nm) = 1242400 = 3.11 eV

E(3,2) = −3.74 − (−5.55) = 1.81 eV ; λ = 12421.81 = 685 nm E(4,3) = −1.57 − (−3.74) = 2.17 eV ; λ = 12422.17 = 572 nm E(4,2) = −1.57 − (−5.55) = 3.99 eV ; λ = 312 nm E(2,1) = −5.55 − (−10.42) = 4.87 eV ; λ = 255.15 nm E(3,1) = −3.74 − (−10.42) = 6.68 eV ; λ = 185.9 nm E(4,1) = −1.57 − (−10.42) = 8.85 eV ; λ = 140.3 nm

E(3,2) = 1.81 · 1.6·10⁻¹⁹ = 2.9·10⁻¹⁹ J E(4,3) = 2.17 · 1.6·10⁻¹⁹ = 3.5·10⁻¹⁹ J

^y₈₃Bi → ²⁰⁶_xTl → ²⁰⁶₈₂Pb
N(t) = N₀·(12)^tT½ ; N(t₁) = N₀16
T½(²³⁴U) = 2.48·10⁵ ; T½(²³⁸U) = 4.5·10⁹ ; t_⊕ = 4.5·10⁹

¹₀n → ¹₁p + ⁰₋₁e + ν̄ A : 1 = 1 + 0 + 0 ; Z : 0 = (+1) + (−1) + 0

¹₁p → ¹₀n + ⁰₊₁e + ν A : 1 = 1 + 0 + 0 ; Z : (+1) = 0 + (+1) + 0

A : 206 = 206 + 0 ; Z : x = 82 + (−1) x = 82 − 1 = 81

²⁰⁶₈₁Tl → ²⁰⁶₈₂Pb + ⁰₋₁e A : 206 = 206 + 0 ✓ ; Z : 81 = 82 + (−1) ✓

Z : 83 = 81 + 2 ✓ ; A : y = 206 + 4 y = 210

²¹⁰₈₃Bi → ²⁰⁶₈₁Tl + ⁴₂He ²¹⁰₈₃Bi —α→ ²⁰⁶₈₁Tl —β⁻→ ²⁰⁶₈₂Pb A : 210 = 206 + 4 + 0 ; Z : 83 = 82 + 2 + (−1) ✓

N(t) = N₀·(12)^tT½ = N₀·2^−tT½

N₀ → N₀2 → N₀4 → N₀8 → N₀16 116 = (12)⁴

N₀16 = N₀·(12)^t₁T½ (12)⁴ = (12)^t₁T½ ⇒ t₁T½ = 4

t₁ = 4·T½

n = tT½ = 4.5·10⁹2.48·10⁵ ≈ 1.8·10⁴

NN₀ = (12)ⁿ = 2⁻¹⁸¹⁴⁵ log₁₀(NN₀) = −n·log₁₀2 = −18145 · 0.301 ≈ −5462 NN₀ ≈ 10⁻⁵⁴⁶²

N = N₀·10⁻⁵⁴⁶² ≈ 10⁸⁰ · 10⁻⁵⁴⁶² = 10⁻⁵³⁸² ≪ 1

n = 4.5·10⁹4.5·10⁹ = 1 ⇒ NN₀ = (12)¹ = 12

²³⁸U —α→ ²³⁴Th —β⁻→ ²³⁴Pa —β⁻→ ²³⁴U

N(²³⁴U)N(²³⁸U) = T½(²³⁴U)T½(²³⁸U) = 2.48·10⁵4.5·10⁹ ≈ 5.5·10⁻⁵