פתרון בגרות · קרינה וחומר

✦ בנוי על ידי פהד גאנם ✦

n | 1u (1/cm) | 1v (1/cm)
1 | 0.033 | 0.009
2 | 0.025 | 0.016
3 | 0.020 | 0.020
4 | 0.017 | 0.025
5 | 0.010 | 0.034
6 | 0.067 | −0.025
7 | 0.083 | −0.047

40.0 cm

24.39 cm

40.0 cm

62.5 cm

−1.56

0.0250

0.0160

0.0410

24.39 cm

4.10 D

1u+1v=1f m = −vu P = 100f D · f in cm

1u + 1v = 1f 1v = 1f − 1u

v₆ = 1−0.025 = −40 cm v₇ = 1−0.047 = −21.3 cm

1u = 1f = 0.041 1/cm ⇒ u = 24.4 cm

1v = −1u + 1f

a = (−0.047) − 0.0340.083 − 0.010 = −0.0810.073 = −1.11

f = 1b

0.042 · 0.041 · 0.040 · 0.042 · 0.044 · 0.042 · 0.036 Σ = 0.287 1/cm 1f = 0.2877 = 0.041 1/cm

f = 10.041 = 24.4 cm

u = v = 2f ⇒ f = 502 = 25 cm

σ√7 = 0.0025√7 = 0.0010 1/cm f = 24.4 ± 0.6 cm

f = 24.4 cm = 0.244 m P = 1f = 10.244 = 4.1 D

P = 0.041 · 100 = 4.1 D

1v = 0.041 − 120 = 0.041 − 0.050 = −0.009 1/cm v = 1−0.009 = −111 cm

m = −vu = −−11120 = +5.6

120 + 1−111.1 = 0.050 − 0.009 = 0.041 = 1f

N = 5300 (1/cm) ; a = 1 m ; L = 1 m
θ (n = 1) : 12.3° , 13.2° , 16.9° , 17.9°
d = 1N ; d · sinθ = n · λ ; 400 nm ≤ λ ≤ 700 nm

843.80 nm

5300 1/cm

1.00 m

0.5000 m

401.94 nm

430.85 nm

548.49 nm

579.92 nm

12.30°

13.20°

16.90°

17.90°

1886.79 nm

0.5000 m

26.57°

843.80 nm

d = 1N d · sinθ = n · λ n_max = dλ tanθ = xL

N = 530010⁻² m = 5.3 · 10⁵ 1/m d = 15.3 · 10⁵ = 1.8868 · 10⁻⁶ m = 1886.8 nm

λ = d · sinθn = d · sinθ

λ₁ = 1886.8 · sin 12.3° = 1886.8 · 0.21303 = 402 nm λ₂ = 1886.8 · sin 13.2° = 1886.8 · 0.22835 = 431 nm λ₃ = 1886.8 · sin 16.9° = 1886.8 · 0.29070 = 548 nm λ₄ = 1886.8 · sin 17.9° = 1886.8 · 0.30736 = 580 nm

λ = sinθN = 0.213035.3 · 10⁵ = 4.0194 · 10⁻⁷ m

sinθ = n · λd ≤ 1 ⇒ n ≤ dλ

n ≤ 1886.8401.9 = 4.694 ⇒ n_max = 4

sinθ₄ = 4 · 401.91886.8 = 0.85212 ⇒ θ₄ = 58.44° sinθ₅ = 5 · 401.91886.8 = 1.0652 > 1

n_max ≤ 1sin 12.3° = 10.21303 = 4.694

d · sinθ = 0 · λ = 0 ⇒ sinθ = 0 ⇒ θ = 0°

sinθ = 4001886.8 = 0.2120 ⇒ θ = 12.24° sinθ = 7001886.8 = 0.3710 ⇒ θ = 21.78°

λ = 1886.8 · sin 12.24° = 400.0 nm λ = 1886.8 · sin 21.78° = 700.0 nm

sinθ = 1 · 7001886.8 = 0.3710 ⇒ θ = 21.78° sinθ = 2 · 4001886.8 = 0.4240 ⇒ θ = 25.09°

(n + 1) · λ_min < n · λ_max 2 · 400 = 800 nm ; 1 · 700 = 700 nm

x = L · tan 21.78° = 0.3995 L ; x = L · tan 25.09° = 0.4682 L

x = a2 = 12 = 0.5 m tanθ = xL = 0.51 = 0.5 ⇒ θ = 26.57°

λ = d · sin 26.57° = 1.8868 · 10⁻⁶ · 0.44721 = 8.438 · 10⁻⁷ m

sinθ = 0.5√0.25 + 1 = 0.51.1180 = 0.44721 = 1√5 λ = d√5 = 843.8 nm

sinθ = 843.81886.8 = 0.4472 ⇒ x = L · tan 26.57° = 0.5 m

n_max ≤ 1886.8843.8 = 2.236 ⇒ n_max = 2 sinθ₂ = 0.8944 ⇒ θ₂ = 63.43° ⇒ x₂ = 2 m

λ = 400 nm ; V₀ = 1.1 V ; hc = 1240 eV·nm
h = 6.626·10⁻³⁴ J·s ; c = 3.00·10⁸ m/s ; e = 1.602·10⁻¹⁹ C
E_k,max = hf − B ; E_k,max = e·V₀

פליטת אלקטרונים · זרם זורם באמפרמטר

400 nm

0.00 V

50 %

7.50 ·10¹⁴ Hz

3.10 eV

2.00 eV

620 nm

1.10 eV

1.10 V

10.0 µA

—

hf = 1240400 = 3.10 eVE_k,max = 3.10 − 2.00 = 1.10 eVV₀ = 1.10 V

E_k,max = hf − B

E_k,max = e·V₀

hf = hcλ = 1240 eV·nm400 nm = 3.10 eV hf = 6.626·10⁻³⁴ · 3.00·10⁸400·10⁻⁹ = 4.9695·10⁻¹⁹ J = 3.10 eV

E_k,max = e·V₀ = 1.1 eV = 1.602·10⁻¹⁹ · 1.1 = 1.7622·10⁻¹⁹ J

B = hf − E_k,max = 3.10 − 1.10 = 2.00 eV B = 4.9695·10⁻¹⁹ − 1.7622·10⁻¹⁹ = 3.21·10⁻¹⁹ J

λ₀ = hcB = 12402.00 = 620 nm E_k,max = 1240 · ( 1400 − 1620 ) = 3.10 − 2.00 = 1.10 eV

E_k = hf − B

hf = 0 ; E_k = 0 − B = −B ; M = ( 0 , −2.00 eV )

E_k = 0 ; hf = B ; N = ( 2.00 eV , 0 ) f₀ = Bh = 3.204·10⁻¹⁹6.626·10⁻³⁴ = 4.84·10¹⁴ Hz λ₀ = cf₀ = 6.20·10⁻⁷ m = 620 nm

x_N = B = 2.00 eV ; |y_M| = |−B| = B = 2.00 eV

ΔE_kΔ(hf) = 1

y_N − y_Mx_N − x_M = 0 − (−B)B − 0 = 1 1 = Bx ; x = B

E_k(2.00) = 2.00 − 2.00 = 0 ; E_k(0) = 0 − 2.00 = −2.00 ; E_k(3.10) = 3.10 − 2.00 = 1.10 eV

0 ≤ hf < 2.00 eV ; λ > 620 nm

E_k = 12 m·v² ≥ 0

n = ∞ : E = 0 eV   |   n = 4 : E = −0.85 eV
n = 3 : E = −1.51 eV   |   n = 2 : E = −3.4 eV
n = 1 : E = −13.6 eV   |   E_k = 12.6 eV
hc = 1240 eV·nm ; h = 6.626·10⁻³⁴ J·s ; c = 3·10⁸ m/s

בחרו אנרגייה ולחצו שגר

12.60 eV

12.60 eV

98.4 nm

n = 3

—

0 מתוך 3

656.1 nm

ΔE_1→3 = (−1.51) − (−13.6) = 12.09 eVλ = 124012.60 = 98.4 nm

ΔE_1→2 = (−3.4) − (−13.6) = 10.20 eV ΔE_1→3 = (−1.51) − (−13.6) = 12.09 eV ΔE_1→4 = (−0.85) − (−13.6) = 12.75 eV ΔE_1→∞ = 0 − (−13.6) = 13.60 eV

12.6 − 12.09 = 0.51 eV

12.6 ≠ 10.20 ; 12.6 ≠ 12.09 ; 12.6 ≠ 12.75

13.6 · ( 1 − 1n² ) = 12.6 ; 13.6n² = 1.0 ; n = 3.69

ΔE_1→2 = 10.20 ≤ 12.6 ; ΔE_1→3 = 12.09 ≤ 12.6 ; ΔE_1→4 = 12.75 > 12.6

12.6 − 12.09 = +0.51 eV ; 12.6 − 12.75 = −0.15 eV

N = n·(n − 1)2 = 3 · 22 = 3

ΔE_3→2 = (−1.51) − (−3.4) = 1.89 eV ; λ = 12401.89 = 656.1 nm ΔE_3→1 = 12.09 eV ; λ = 124012.09 = 102.6 nm ΔE_2→1 = 10.20 eV ; λ = 124010.20 = 121.6 nm

1λ = R · ( 14 − 19 ) = 1.524·10⁶ 1/m ; λ = 656.3 nm

E_min = 0 − (−13.6) = 13.6 eV

λ_max = hcE_min = 124013.6 = 91.2 nm

E = 13.6 · 1.6·10⁻¹⁹ = 2.176·10⁻¹⁸ J λ = 6.626·10⁻³⁴ · 3·10⁸2.176·10⁻¹⁸ = 9.14·10⁻⁸ m = 91.4 nm

λ < 91.2 nm ; E = 1240λ > 13.6 eV

E_k = 1240λ − 13.6 eV

124080 = 15.5 eV ; E_k = 15.5 − 13.6 = 1.9 eV

R₀ = 2.336 · 10⁸ Bq ; t : 0 → 264 h ; Δt = 11 h
Mo : A = 99 , Z = 42 ; Tc : A = 99 , Z = 43 ; T½(Tc) ≫ T½(Mo)
N_Mo(t) + N_Tc(t) = N₀ = const

0.00 h

×5

2.336·10⁸

0.00

0.0000

8.007·10¹³

2.336·10⁸

100.00 %

λ = ln 2T½ = 0.6931472237600 s = 2.9173·10⁻⁶ s⁻¹ N₀ = R₀λ = 8.007·10¹³ N_Mo(t) = N₀ · e^−λ·t

Mo-99 : Z = 42 p , N = 57 n , A = Z + N = 99

10n → 11p + 0−1e + 00ν̄

0 = (+1) + (−1) + 0 = 0 1 = 1 + 0 + 0 = 1 0 = 0 + (+1) + (−1) = 0

9942Mo → 9943Tc + 0−1e + 00ν̄

A : 99 = 99 + 0 + 0 = 99 Z : 42 = 43 + (−1) + 0 = 42

Mo : 42 p , 57 n Tc : 43 p , 56 n

t = 6 · 11 = 66 h

N_Mo = N_Tc N_Tc = N₀ − N_Mo

N_Mo = N₀ − N_Mo ⟹ 2N_Mo = N₀ ⟹ N_Mo = N₀2

N(132) = N₀4 = 2.002·10¹³ 26466 = 4 ; N(264) = N₀16 = 5.005·10¹²

T½ = 66 · 3600 = 237600 s λ = ln 2T½ = 0.6931472237600 = 2.9173·10⁻⁶ s⁻¹

R = λ · N ⟹ N₀ = R₀λ = 2.336·10⁸2.9173·10⁻⁶ = 8.007·10¹³

n = tT½ = 16866 = 2.5455 N = N₀ · 12ⁿ = 8.007·10¹³ · 2^−2.5455 = 8.007·10¹³ · 0.17129

t = 168 · 3600 = 604800 s ⟹ λ·t = 1.76437 N = N₀ · e^−λ·t = 8.007·10¹³ · 0.17129 = 1.372·10¹³ R(168) = R₀ · e^−λ·t = 4.001·10⁷ Bq N = Rλ = 4.001·10⁷2.9173·10⁻⁶ = 1.372·10¹³

n_Tc = 143 hT½(Tc) ≈ 0 ⟹ 2⁻ⁿ ≈ 1

N_Mo(t) + N_Tc(t) = N₀ = 8.007·10¹³

n = 14366 = 2.1667 ⟹ 2^−2.1667 = 0.22272 N_Mo(143) = 8.007·10¹³ · 0.22272 = 1.783·10¹³ N_Tc(143) = N₀ − N_Mo = 6.224·10¹³ 1.783·10¹³ + 6.224·10¹³ = 8.007·10¹³ = N₀

E_γ = h · f = E_מעורר − E_יסוד

AZX* → AZX + γ

Mo-99 (Z = 42) → Tc-99 (Z = 43) + e⁻ + ν̄

T½(Mo) = 66 h = 237600 s λ = 2.9173·10⁻⁶ s⁻¹ N₀ = 8.007·10¹³ N(168 h) = 1.372·10¹³ N_Mo(143) + N_Tc(143) = 8.007·10¹³