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| '''Calculation of Gamow Factor for Alpha decay of Nuclei'''
| | At the turning point, |
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| [[Image:Gamow.jpg]]
| | <math>E=V(b)=\frac{2z_{1}e^{2}}{b},</math> |
| Since the <math>\alpha</math>-decay happens in the nulcie then we can assume that an <math>\alpha</math>-decay is formed in the nucleus just before its emission (although <math>\alpha</math> particle doesnot exist in the nucleus).
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| Inside the nucleus the particle will experience nuclear interaction which mostly attractive and outside the nucleus the inetraction would be coulombic(replusive).
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| Since the mathematical form of the nuclear interaction is not known we can model it by a square well type potential for the present purpose.
| | so that |
| Outside the range of the nuclear interaction would be coulombic. So the coulomb interaction is,
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| <math>V_{coul} = \frac{1}{4\pi\epsilon_{0}}\frac{2z_{1}e^{2}}{r}</math> | | <math>b=\frac{2z_{1}e^{2}}{E}.</math> |
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| where <math>Z_{1}</math> is the atomic number of the rest of the nucleus(after decay).
| | Within the WKB approximation, the transmission probability is given by |
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| From the WKB apporximation we know that at the turning point, <math>E= V(x)= V_{coul} = \frac{1}{4\pi\epsilon_{0}}\frac{2z_{1}e^{2}}{R_{c}}</math>
| | <math>T=\exp\left [-2\int_{a}^{b}p(x)\,dx\right ],</math> |
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| <math>R_{c} = \frac{1}{4\pi\epsilon_{0}}\frac{2z_{1}e^{2}}{E}</math> | | where <math>p(x)=\frac{1}{\hbar}\sqrt{2m\left(V(x)-E\right)}.</math> |
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| Now the Transition probabilty
| | We now evaluate the integral appearing in the exponential. |
| <math>T\cong \Theta ^{2}</math>, | | |
| where <math>\Theta = e^{-\int_{b}^{a}q(x)dx}</math>
| | <math>\int_{a}^{b}p(x)\,dx=\sqrt{\frac{2m}{\hbar^2}}\int_{a}^{b}\sqrt{V(x)-E}\,dx = \sqrt{\frac{2m}{\hbar^{2}}}\int_{a}^{b} |
| | \sqrt{\frac{2z_{1}e^{2}}{x}-E}\,dx</math> |
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| and <math>q(x)= \frac{1}{\hbar}\sqrt{2m\left(V(x)-E\right)}</math>
| | <math>=\sqrt{\frac{4mz_{1}e^{2}}{\hbar^2}}\int_{a}^{b} \sqrt{\frac{1}{x}-\frac{1}{b}}\,dx</math> |
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| <math>\Theta ^{2} = e^{-2\int_{b}^{a} q(x)dx}</math>
| | Let us define |
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| In the present problem <math>b= R</math> and <math>a = R_{c}</math>
| | <math>I=\int_{a}^{b} \sqrt{\frac{1}{x}-\frac{1}{b}}\,dx.</math> |
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| Now,
| | We now make the substitution, |
| <math>\int_{R}^{R_{c}} \left(\frac{2m}{\hbar^{2}}\right)^{\frac{1}{2}}(V(x)-E)^{\frac{1}{2}} dr = \left(\frac{2m}{\hbar^{2}}\right)^{\frac{1}{2}}\int_{R}^{R_{c}} | | |
| \left(\frac{1}{4\pi\epsilon_{0}}\frac{2z_{1}e^{2}}{r}-E\right)^\frac{1}{2}dr</math>
| | <math>x=b\cos^{2}\theta.\!</math> |
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| <math>= \left(\frac{2m}{\hbar^{2}}\right)^{\frac{1}{2}}\left(\frac{2z_{1}e^{2}}{4\pi\epsilon_{0}}\right)^{\frac{1}{2}}\int_{R}^{R_{c}} \left [ \frac{1}{r} - \frac{1}{R_{c}}\right ]^{\frac{1}{2}}dr</math>
| | We then obtain |
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| let, <math>I = \int_{R}^{R_{c}} \left [ \frac{1}{r} - \frac{1}{R_{c}}\right ]^{\frac{1}{2}}dr</math>
| | <math>I= 2\int_{0}^{\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )} \sqrt{\frac{b\sin^{2}\theta}{\cos^{2}\theta}}\cos\theta\sin\theta\, d\theta</math> |
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| | <math>=2\sqrt{b}\int_{0}^{\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )}\sin^{2}\theta\,d\theta </math> |
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| Put,
| | <math>=\sqrt{b}\int_{0}^{\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )} ( 1-\cos{2\theta})\,d\theta</math> |
| <math>r= R_{0}cos^{2}\theta</math> | |
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| <math>dr= -R_{0}2cos\theta sin\theta</math> | | <math>=\sqrt{b}\left [ \theta - \sin\theta \cos\theta \right ]_{0}^{\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )}</math> |
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| <math>I= 2\int_{0}^{cos^{-1}\sqrt{\frac{R}{R_{c}}}} \left( \frac{R_{c}sin^{2}\theta}{cos^{2}\theta}\right)^{\frac{1}{2}} cos\theta sin\theta d\theta</math> | | <math>=\sqrt{b}\left \{\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )-\sin\left [\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )\right ]\cos\left [\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )\right ]\right \}</math> |
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| <math>2R_{c}^{\frac{1}{2}}\int_{0}^{cos^{-1}\sqrt{\frac{R}{R_{c}}}} sin^{2}\theta d\theta </math> | | <math>=\sqrt{b}\left [\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )-\sqrt{\frac{a}{b}}\sqrt{1-\frac{a}{b}} \right ]</math> |
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| <math>I= R_{c}^{\frac{1}{2}}\int_{0}^{cos^{-1}\sqrt{\frac{R}{R_{c}}}} ( 1-{cos2\theta}) d\theta</math> | | Let us consider the limit, <math>b\gg a.</math> We then have |
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| <math>I= R_{c}^{\frac{1}{2}}\left [ \theta - sin\theta cos\theta \right ]_{0}^{cos^{-1}\sqrt{\frac{R}{R_{c}}}}</math> | | <math>I\approx\frac{\pi}{2}\sqrt{b}-2\sqrt{a},</math> |
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| <math>I= R_{c}^{\frac{1}{2}}\left [ cos^{-1}\sqrt{\frac{R}{R_{c}}} - sin \left(cos^{-1}\sqrt{\frac{R}{R_{c}}}\right) cos\left(cos^{-1}\sqrt{\frac{R}{R_{c}}}\right) \right ]</math> | | where we use the fact that <math>\cos^{-1}{x}\approx\frac{\pi}{2}-x.</math> |
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| <math>I= R_{c}^{\frac{1}{2}}\left [ cos^{-1}\sqrt{\frac{R}{R_{c}}} - \sqrt{\frac{R}{R_{c}}}\sqrt{1- \frac{R}{R_{c}}} \right ]</math>
| | Combining all of the above results, we get |
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| <math>I= R_{c}^{\frac{1}{2}}\left [ cos^{-1}\sqrt{\frac{R}{R_{c}}} - \sqrt{\frac{R}{R_{c}}- \left(\frac{R}{R_{c}}\right)^{2}} \right ]</math> | | <math> T=\exp\left (-\frac{2\pi z_{1}e^{2}}{\hbar}\sqrt{\frac{2m}{E}}+\frac{4}{\hbar}\sqrt{4mz_{1}e^{2}a}\right ).</math> |
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| Let us consider <math>R_{c} \gg R</math>
| | We may express this in terms of the velocity of the alpha particle by noting that the kinetic energy <math>E=\tfrac{1}{2}mv^{2}.</math> Doing so, we obtain |
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| Then we have
| | <math> T=\exp\left (-\frac{4\pi z_{1}e^{2}}{\hbar v} \right )\exp\left (\frac{8e}{\hbar}\sqrt{z_{1}ma}\right ).</math> |
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| <math>I\cong \sqrt{R_{c}}\left(cos^{-1}\sqrt{\frac{R}{R_{c}}}-\sqrt{\frac{R}{R_{c}}} \right)</math>
| | The first exponential factor is known as the Gamow factor. The Gamow factor determines the dependence of the transmission probability on the speed (or energy) of the alpha particle. |
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| where <math>cos^{-1}\sqrt{\frac{R}{R_{c}}} \cong \frac{\pi}{2} - \left(\frac{R}{R_{c}}\right)^{\frac{1}{2}}</math>
| | Back to [[WKB Approximation#Problems|WKB Approximation]] |
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| Setting, charge of <math>\alpha</math>particle = 2= <math>Z_{2}</math>(in general)
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| <math>\int q(x)dx = \left ( \frac{2Mz_{1}z_{2}e^{2}R_{c}}{\hbar^{2}4\pi\epsilon_0} \right )^{\frac{1}{2}}\left [\frac{\pi}{2} - 2\left(\frac{R}{R_{c}}\right)^{\frac{1}{2}} \right ]</math>
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| Now <math> T\cong e^{-2\int q(x)dx} = exp\left [ -\frac{\pi z_{1}z_{2}e^{2}}{\hbar 4\pi\epsilon_0} \left (\frac{2M}{e} \right )^{2} + \frac{4}{\hbar} \left ( \frac{2z_{1}z_{2}e^{2}MR}{4\pi\epsilon_0} \right )^{\frac{1}{2}}\right ]</math>
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| Now putting <math>E= \frac{1}{2}mv^{2}</math>, veloctiy of the particle
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| <math> T\cong exp\left ( \frac{-2\pi z_{1}z_{2}e^{2}}{4\pi\epsilon_0\hbar v} \right )exp \left ( \frac{32z_{1}z_{2}e^{2}MR}{4\pi\epsilon_0\hbar^{2} } \right )^{\frac{1}{2}}</math>
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| The 1st exponential term is known as the Gamow factor. The Gamow factor determines the dependence of the probability on the speed (or energy) of the alpha particle.
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At the turning point,
so that
Within the WKB approximation, the transmission probability is given by
![{\displaystyle T=\exp \left[-2\int _{a}^{b}p(x)\,dx\right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/362117ffb4750dda048f6b9fe4014da8b5a16ece)
where 
We now evaluate the integral appearing in the exponential.
Let us define
We now make the substitution,
We then obtain
![{\displaystyle ={\sqrt {b}}\left[\theta -\sin \theta \cos \theta \right]_{0}^{\cos ^{-1}\left({\sqrt {\frac {a}{b}}}\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea11402c84a2ce79e18a337ed3577fb1fb9a2342)
![{\displaystyle ={\sqrt {b}}\left\{\cos ^{-1}\left({\sqrt {\frac {a}{b}}}\right)-\sin \left[\cos ^{-1}\left({\sqrt {\frac {a}{b}}}\right)\right]\cos \left[\cos ^{-1}\left({\sqrt {\frac {a}{b}}}\right)\right]\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/90d94364ad0669a2c34a85906705fa7648ea2cf6)
![{\displaystyle ={\sqrt {b}}\left[\cos ^{-1}\left({\sqrt {\frac {a}{b}}}\right)-{\sqrt {\frac {a}{b}}}{\sqrt {1-{\frac {a}{b}}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c2dd00248a17116c3a1ad2eaa53021ef8288100)
Let us consider the limit, 
We then have
where we use the fact that 
Combining all of the above results, we get
We may express this in terms of the velocity of the alpha particle by noting that the kinetic energy 
Doing so, we obtain
The first exponential factor is known as the Gamow factor. The Gamow factor determines the dependence of the transmission probability on the speed (or energy) of the alpha particle.
Back to WKB Approximation