פתרון בגרות · קרינה וחומר
✦ בנוי על ידי פהד גאנם ✦
العربية
☀ מצב בהיר
1
▾
✕
1
2.00
12.5
25
0
λ =
v
f
L =
λ
2
→ λ = 2L = 2 · 1 = 2 m
v = λ · f = 2 · 12.5 = 25 m/s
v =
√
T
μ
v = λ · f =
קבוע
→ f ↑ ⇒ λ ↓
L = 3 ·
λ₃
2
→ λ₃ =
2L
3
=
2
3
≈ 0.667 m
f₃ =
v
λ₃
=
25
0.667
= 37.5 Hz
f
n
= n ·
v
2L
→ f₃ = 3 · 12.5 = 37.5 Hz
L = n ·
λ
2
, n = 1, 2, 3, …
λ =
v
f
=
25
60
≈ 0.417 m
n =
2L
λ
=
2 · 1
0.417
= 4.8
f
n
= n · f₁ = n · 12.5 →
60
12.5
= 4.8
v =
√
T
μ
→ v
א
= v
ב
f₁ =
v
2L
f
א
> f
ב
⇒
v
2L
א
>
v
2L
ב
⇒ L
א
< L
ב
2
▾
✕
d · sin θ
m
= m · λ · y
m
= L · tan θ
m
· λ =
c
f
· m ≤
d
λ
1.20 cm
5.0 cm
1.20
5.0
9
24.75
v =
c
n
, n ≥ 1
λ =
v
f
→ λ
n
=
λ₀
n
λ =
c
f
λ
min
=
c
f
max
=
3·10⁸
300·10⁹
=
3·10⁸
3·10¹¹
= 1·10⁻³ m
λ
min
= 1·10⁻³ m = 1 mm
λ
max
=
3·10⁸
3·10⁸
= 1 m
r₁ = r₂ → Δr = r₂ − r₁ = 0
Δr = m · λ , m = 0 → Δr = 0
d · sin θ₁ = 1 · λ
tan θ₁ =
y₁
L
=
0.2475
1
= 0.2475 → θ₁ = 13.90°
sin θ₁ =
y₁
√
y₁² + L²
=
0.2475
1.0302
= 0.24025
λ =
d · sin θ₁
m
=
0.05 · 0.24025
1
= 1.201·10⁻² m ≈ 1.2 cm
λ
ap
=
d · y₁
L
=
0.05 · 0.2475
1
= 1.2375·10⁻² m ≈ 1.24 cm
sin θ
m
=
m · λ
d
≤ 1 → m ≤
d
λ
=
5
1.2
= 4.17
m = 1 → sin θ = 0.240 ✓
m = 2 → sin θ = 0.481 ✓
m = 3 → sin θ = 0.721 ✓
m = 4 → sin θ = 0.961 ✓
m = 5 → sin θ = 1.201 ✕
N = 2 · m
max
+ 1 = 2 · 4 + 1 = 9
θ₄ = 73.9° → y₄ = L · tan θ₄ = 1 · 3.46 = 3.46 m
3
▾
✕
400 nm
400
2.500
3.00
1.00
1.00
h·f = B + E
k,max
E
k,max
= e · V
f =
c
λ
e · V =
h · c
λ
− B
V =
h · c
e · λ
−
B
e
V =
h · c
e
·
1
λ
−
B
e
V = 0 → E
k,max
= e · V = 0 → h·f₀ = B
1
λ₀
= 1.667 · 10⁶ m⁻¹ → λ₀ =
1
1.667 · 10⁶
= 6.0 · 10⁻⁷ m = 600 nm
f₀ =
c
λ₀
=
3 · 10⁸
6.0 · 10⁻⁷
= 5 · 10¹⁴ Hz
a =
ΔV
Δ
1
λ
=
1.00 − 0.40
(2.5 − 2.0) · 10⁶
=
0.60
0.5 · 10⁶
= 1.2 · 10⁻⁶ V·m
a =
h · c
e
→ h =
a · e
c
h =
1.2 · 10⁻⁶ · 1.6 · 10⁻¹⁹
3 · 10⁸
= 6.4 · 10⁻³⁴ J·s
0 =
h · c
e · λ₀
−
B
e
→ B =
h · c
λ₀
B = a · e ·
1
λ₀
= 1.2 · 10⁻⁶ · 1.6 · 10⁻¹⁹ · 1.667 · 10⁶ = 3.2 · 10⁻¹⁹ J
B =
3.2 · 10⁻¹⁹
1.6 · 10⁻¹⁹
= 2.0 eV
B[eV] = 1.2 · 10⁻⁶ · 1.667 · 10⁶ = 2.0 eV
1
λ
=
1
400 · 10⁻⁹
= 2.5 · 10⁶ m⁻¹
V = a ·
1
λ
− B[eV] = 1.2 · 10⁻⁶ · 2.5 · 10⁶ − 2.0 = 3.0 − 2.0 = 1.0 V
E
photon
= 3.0 eV , B = 2.0 eV , E
k,max
= 1.0 eV
E
photon
= h·f , h·f = B + e·V
4
▾
✕
n = ∞ E = 0.0 eV
n = 4 E = −1.6 eV
n = 3 E = −3.7 eV
n = 2 E = −5.5 eV
n = 1 E = −10.4 eV
7.0 V
185 nm
—
7.0
—
—
—
E
photon
= E
high
− E
low
= h · f
λ =
h · c
ΔE
E
k
= q · V = e · 7 V = 7 eV
E
k
≥ ΔE
excitation
E₂ − E₁ = (−5.5) − (−10.4) = 4.9 eV ≤ 7 eV
E₃ − E₁ = (−3.7) − (−10.4) = 6.7 eV ≤ 7 eV
E₄ − E₁ = (−1.6) − (−10.4) = 8.8 eV > 7 eV
3 → 2 : ΔE = (−3.7) − (−5.5) = 1.8 eV
3 → 1 : ΔE = (−3.7) − (−10.4) = 6.7 eV
2 → 1 : ΔE = (−5.5) − (−10.4) = 4.9 eV
ΔE = h · f =
h · c
λ
→ λ =
h · c
ΔE
ΔE
min
= E₃ − E₂ = (−3.7) − (−5.5) = 1.8 eV
λ
max
=
h · c
ΔE
min
=
1240 eV·nm
1.8 eV
= 689 nm
λ
max
=
6.626·10⁻³⁴ · 3·10⁸
1.8 · 1.6·10⁻¹⁹
= 6.9·10⁻⁷ m
4.9 eV · 6.7 eV · 8.8 eV · E
ion
= 10.4 eV
E₁ =
h · c
λ₁
=
1240 eV·nm
185 nm
= 6.70 eV
6.70 eV = E₃ − E₁ → n = 1 → 3
E₂ =
1240 eV·nm
170 nm
= 7.29 eV
4.9 < 7.29 < 8.8 < 10.4
E
photon
=
h · c
λ
=
1240 eV·nm
82 nm
= 15.1 eV
E
ion
= 0 − (−10.4) = 10.4 eV
E
k
= E
photon
− E
ion
= 15.1 − 10.4 = 4.7 eV
E
k
= 4.7 · 1.6·10⁻¹⁹ = 7.6·10⁻¹⁹ J
5
▾
✕
1
8
0
2
6
3.5
3
5
5.4
4
2
16.0
5
0.5
32.0
0.0 d
0.0
8.00
1.000
0.00
1.000
F = q·v·B·sinθ
q
α
= +2e , q
β
= −e , q
γ
= 0
q
γ
= 0 → F = 0 →
no deflection
A
Z
X
*
→
A
Z
X + γ
E
γ
= h·f = E
2
− E
1
A = A + 0 , Z = Z + 0
N(t) = N₀·e
−λt
N: 8 → 4 , t: 0 → 8 ⇒ T½ = 8 d
T½ =
16
2
= 8 d , T½ =
32
4
= 8 d
λ =
ln 2
T½
=
0.693
8
= 0.0866
day⁻¹
T½ = 8 · 24 · 3600 = 6.912·10⁵ s
λ =
0.693
6.912·10⁵
= 1.0·10⁻⁶
s⁻¹
R = λ·N ⇒
R
10
= λ·
N
10
N
N₀
= e
−λt
=
1
10
λ·t = ln 10 ⇒ t =
ln 10
λ
=
2.303
0.0866
= 26.6 d
n = log₂10 = 3.32 ⇒ t = n·T½ = 3.32 · 8 = 26.6 d
N₀
2
= N₀·e
−λT½
⇒ T½ =
ln 2
λ
2N₀ · e
−λ·8
= 2N₀ ·
1
2
= N₀
→
🏠
♿