Gamov nazariyasi
Gamov nazariyasi - radioaktiv yadrolarning
α
{\displaystyle \alpha }
-parchalanishini kvant tasavvurlar asosida tushuntirib beruvchi nazariya.
α
{\displaystyle \alpha }
-parchalanish deganda, ogʻir yadrolarning oʻzidan geliy atomining yadrosini chiqarib, boshqa yadroga aylanishi tushuniladi.
α
{\displaystyle \alpha }
-parchalanish uchun energetik shart quyidagi koʻrinishga ega:
M
(
A
,
Z
)
>
M
(
A
−
4
,
Z
−
2
)
+
M
α
{\displaystyle M(A,\ Z)>M(A-4,\ Z-2)+M_{\alpha }}
Parchalanishda hosil boʻlgan energiya quyidagi ifoda orqali topiladi:
Q
α
=
(
M
(
A
,
Z
)
−
M
(
A
−
4
,
Z
−
2
)
−
M
α
)
c
2
,
{\displaystyle Q_{\alpha }=(M(A,\ Z)-M(A-4,\ Z-2)-M_{\alpha })c^{2},}
Energiya va impulsning saqlanish qonunlaridan foydalanib,
α
{\displaystyle \alpha }
-zarra yadroni tark etayotganida ega boʻladigan kinetik energiyasini aniqlash mumkin:
T
α
=
M
(
A
−
4
,
Z
−
2
)
m
α
+
M
(
A
−
4
,
Z
−
2
)
Q
α
{\displaystyle T_{\alpha }={\dfrac {M(A-4,\ Z-2)}{m_{\alpha }+M(A-4,\ Z-2)}}Q_{\alpha }}
α
{\displaystyle \alpha }
-parchalanishda hosil boʻlgan
α
{\displaystyle \alpha }
-zarra kinetik energiyasi 2-8 MeV atrofida boʻladi,
α
{\displaystyle \alpha }
-zarraning yadroga bogʻlanish energiyasi esa 28.3 MeV ga teng. U holda
α
{\displaystyle \alpha }
-zarra yadroni qanday tark etadi, degan tabiiy savol tugʻiladi. Bunga G. A. Gamov kvant tasavvurlar asosida ishlab chiqqan nazariyasi orqali javob topish mumkin.
Gamov nazariyasi quyidagi prinsiplarga asoslanadi:
Vaqt birligidagi parchalanish ehtimolligi
λ
{\displaystyle \lambda }
ni quyidagi ifoda orqali aniqlash mumkin:
λ
=
n
T
{\displaystyle \lambda =nT}
n
{\displaystyle n}
—
α
{\displaystyle \alpha }
-zarraning yadro ichida potensial toʻsiq bilan toʻqnashishlari soni,
T
{\displaystyle T}
—
α
{\displaystyle \alpha }
-zarraning toʻsiq orqali oʻtish ehtimolligi.
Aytaylik, ixtiyoriy vaqt momentida yadro ichida faqat bitta
α
{\displaystyle \alpha }
-zarra mavjud boʻlsin va u yadro diametri boʻylab oldinga va orqaga harakatlanayotgan boʻlsin.
ν
=
v
2
R
0
{\displaystyle \nu ={\dfrac {v}{2R_{0}}}}
v
{\displaystyle v}
—
α
{\displaystyle \alpha }
-zarraning yadroni tark etayotgandagi tezligi.
Kengligi
L
{\displaystyle L}
boʻlgan toʻsiqdan zarraning oʻtish ehtimolligi:
T
=
e
−
2
k
2
L
,
(
1
)
{\displaystyle T=e^{-2k_{2}L},\ \ \ \ \ \ \ \ \ (1)}
k
2
=
2
m
(
U
−
E
)
ℏ
{\displaystyle k_{2}={\dfrac {\sqrt {2m(U-E)}}{\hbar }}}
(1) tenglama toʻgʻri burchakli potensial toʻsiqni ifodalaydi,
α
{\displaystyle \alpha }
-zarra yadro ichida toʻsiq bilan koʻp marta toʻqnashadi.
ln
T
=
−
2
k
2
L
,
(
2
)
{\displaystyle \ln T=-2k_{2}L,\ \ \ \ \ \ \ \ (2)}
ln
T
=
−
2
∫
0
L
k
2
(
r
)
d
r
=
−
2
∫
R
0
R
k
2
(
r
)
d
r
,
(
3
)
{\displaystyle \ln T=-2\int \limits _{0}^{L}k_{2}(r)\,dr=-2\int \limits _{R_{0}}^{R}k_{2}(r)\,dr,\ \ \ \ \ \ \ \ \ (3)}
R
0
{\displaystyle R_{0}}
— yadro radiusi,
R
{\displaystyle R}
—
U
=
E
{\displaystyle U=E}
boʻlganda, yadrogacha masofa
α
{\displaystyle \alpha }
-zarraning
r
{\displaystyle r}
masofadagi elektr potensial energiyasi:
U
(
r
)
=
2
Z
e
2
4
π
ε
0
r
{\displaystyle U(r)={\dfrac {2Ze^{2}}{4\pi \varepsilon _{0}r}}}
U holda,
k
2
=
2
m
(
U
−
E
)
ℏ
=
(
2
m
ℏ
2
)
1
/
2
⋅
(
2
Z
e
2
4
π
ε
0
r
−
E
)
1
/
2
{\displaystyle k_{2}={\dfrac {\sqrt {2m(U-E)}}{\hbar }}=\left({\dfrac {2m}{\hbar ^{2}}}\right)^{1/2}\cdot \left({\dfrac {2Ze^{2}}{4\pi \varepsilon _{0}r}}-E\right)^{1/2}}
r
=
R
{\displaystyle r=R}
boʻlganda
U
=
E
{\displaystyle U=E}
boʻlgani uchun,
E
=
2
Z
e
2
4
π
ε
0
R
{\displaystyle E={\dfrac {2Ze^{2}}{4\pi \varepsilon _{0}R}}}
k
2
{\displaystyle k_{2}}
ni quyidagicha yozish mumkin:
k
2
=
(
2
m
ℏ
2
)
1
/
2
⋅
(
R
r
−
1
)
1
/
2
{\displaystyle k_{2}=\left({\dfrac {2m}{\hbar ^{2}}}\right)^{1/2}\cdot \left({\dfrac {R}{r}}-1\right)^{1/2}}
ln
T
=
−
2
∫
R
0
R
k
2
(
r
)
d
r
=
−
2
(
2
m
E
ℏ
2
)
1
/
2
⋅
(
R
r
−
1
)
1
/
2
d
r
(
4
)
{\displaystyle \ln T=-2\int \limits _{R_{0}}^{R}k_{2}(r)\,dr=-2\left({\dfrac {2mE}{\hbar ^{2}}}\right)^{1/2}\cdot \left({\dfrac {R}{r}}-1\right)^{1/2}\,dr\ \ \ \ \ \ \ \ (4)}
[
r
=
R
cos
2
θ
,
d
r
=
−
2
R
sin
θ
cos
θ
d
θ
]
{\displaystyle \left[r=R\cos ^{2}\theta ,\ \ \ dr=-2R\sin \theta \cos \theta d\theta \right]}
[
∫
(
R
r
−
1
)
1
/
2
d
r
=
−
2
R
∫
sin
2
θ
d
θ
]
{\displaystyle \left[\int \left({\dfrac {R}{r}}-1\right)^{1/2}\,dr=-2R\int \sin ^{2}\theta \,d\theta \right]}
ln
T
=
−
2
(
2
m
E
ℏ
2
)
1
/
2
R
[
cos
−
1
(
R
0
R
)
1
/
2
−
(
R
0
R
)
1
/
2
(
1
−
R
0
R
)
1
/
2
]
{\displaystyle \ln T=-2\left({\dfrac {2mE}{\hbar ^{2}}}\right)^{1/2}R\left[\cos ^{-1}\left({\dfrac {R_{0}}{R}}\right)^{1/2}-\left({\dfrac {R_{0}}{R}}\right)^{1/2}\left(1-{\dfrac {R_{0}}{R}}\right)^{1/2}\right]}
Potensial toʻsiq yetarlicha keng boʻlgani uchun,
R
≫
R
0
{\displaystyle R\gg R_{0}}
hamda
cos
(
π
2
−
θ
)
=
sin
θ
{\displaystyle \cos \left({\dfrac {\pi }{2}}-\theta \right)=\sin \theta }
sin
(
R
0
R
)
1
/
2
≈
(
R
0
R
)
1
/
2
{\displaystyle \sin \left({\dfrac {R_{0}}{R}}\right)^{1/2}\approx \left({\dfrac {R_{0}}{R}}\right)^{1/2}}
cos
−
1
(
R
0
R
)
1
/
2
≈
π
2
−
(
R
0
R
)
1
/
2
{\displaystyle \cos ^{-1}\left({\dfrac {R_{0}}{R}}\right)^{1/2}\approx {\dfrac {\pi }{2}}-\left({\dfrac {R_{0}}{R}}\right)^{1/2}}
1
−
(
R
0
R
)
1
/
2
≈
1
{\displaystyle 1-\left({\dfrac {R_{0}}{R}}\right)^{1/2}\approx 1}
ln
T
=
−
2
(
2
m
E
ℏ
2
)
1
/
2
r
[
π
2
−
2
(
R
0
R
)
1
/
2
]
{\displaystyle \ln T=-2\left({\dfrac {2mE}{\hbar ^{2}}}\right)^{1/2}r\left[{\dfrac {\pi }{2}}-2\left({\dfrac {R_{0}}{R}}\right)^{1/2}\right]}
[
R
=
2
Z
e
2
4
π
ε
0
E
]
{\displaystyle \left[R={\dfrac {2Ze^{2}}{4\pi \varepsilon _{0}E}}\right]}
Bundan kelib chiqadiki,
ln
T
=
4
e
ℏ
(
m
π
ε
0
)
1
/
2
Z
1
/
2
R
0
1
/
2
−
e
2
ℏ
ε
0
(
m
2
)
1
/
2
Z
E
−
1
/
2
{\displaystyle \ln T={\dfrac {4e}{\hbar }}\left({\dfrac {m}{\pi \varepsilon _{0}}}\right)^{1/2}Z^{1/2}R_{0}^{1/2}-{\dfrac {e^{2}}{\hbar \varepsilon _{0}}}\left({\dfrac {m}{2}}\right)^{1/2}ZE^{-1/2}}
Tenglamadagi doimiylarning qiymatlarini oʻrniga qoʻyib hisoblasak:
ln
T
=
2
,
97
Z
1
/
2
R
0
1
/
2
−
3
,
95
Z
E
−
1
/
2
{\displaystyle \ln T=2,97Z^{1/2}R_{0}^{1/2}-3,95ZE^{-1/2}}
E
{\displaystyle E}
— energiya,
R
0
{\displaystyle R_{0}}
— yadro radiusi,
Z
{\displaystyle Z}
— hosilaviy yadroning tartib raqami.
log
10
A
=
(
log
10
e
)
(
ln
A
)
=
0
,
4343
ln
A
{\displaystyle \log _{10}A=\left(\log _{10}e\right)(\ln A)=0,4343\ln A}
log
10
T
=
1
,
29
Z
1
/
2
R
0
1
/
2
−
1
,
72
Z
E
−
1
/
2
{\displaystyle \log _{10}T=1,29Z^{1/2}R_{0}^{1/2}-1,72ZE^{-1/2}}
Taʼrifga binoan, parchalanish doimiysi
λ
=
ν
T
=
v
2
R
0
T
{\displaystyle \lambda =\nu T={\dfrac {v}{2R_{0}}}T}
Tenglamaning ikkala tomonidan
log
10
{\displaystyle \log _{10}}
olamiz va
T
{\displaystyle T}
bilan almashtiramiz:
log
10
λ
=
log
10
(
v
2
R
0
)
+
1
,
92
Z
1
/
2
R
0
1
/
2
−
1
,
72
Z
E
−
1
/
2
{\displaystyle \log _{10}\lambda =\log _{10}\left({\dfrac {v}{2R_{0}}}\right)+1,92Z^{1/2}R_{0}^{1/2}-1,72ZE^{-1/2}}
Hosil boʻlgan bu formulaga Geyger-Nettol qonuni deyiladi. Ushbu qonun
α
{\displaystyle \alpha }
-parchalanish energiyasi va radioaktiv yadrolarning yarim yemirilish davrlari orasidagi bogʻliqlikni ifodalaydi va katta amaliy ahamiyatga ega.
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