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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>
| | At the turning point, |
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| <math>R_{c} = \frac{1}{4\pi\epsilon_{0}}\frac{2z_{1}e^{2}}{E}</math> | | <math>E=V(b)=\frac{2z_{1}e^{2}}{b},</math> |
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| Now the Transition probabilty
| | so that |
| <math>T\cong \Theta ^{2}</math>,
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| where <math>\Theta = e^{-\int_{b}^{a}q(x)dx}</math>
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| and <math>q(x)= \frac{1}{\hbar}\sqrt{2m\left(V(x)-E\right)}</math>
| | <math>b=\frac{2z_{1}e^{2}}{E}.</math> |
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| <math>\Theta ^{2} = e^{-2\int_{b}^{a} q(x)dx}</math>
| | Within the WKB approximation, the transmission probability is given by |
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| In the present problem <math>b= R</math> and <math>a = R_{c}</math>
| | <math>T=\exp\left [-2\int_{a}^{b}p(x)\,dx\right ],</math> |
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| Now,
| | where <math>p(x)=\frac{1}{\hbar}\sqrt{2m\left(V(x)-E\right)}.</math> |
| <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>
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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 now evaluate the integral appearing in the exponential. |
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| | <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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| let, <math>I = \int_{R}^{R_{c}} \left [ \frac{1}{r} - \frac{1}{R_{c}}\right ]^{\frac{1}{2}}dr</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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| | Let us define |
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| Put,
| | <math>I=\int_{a}^{b} \sqrt{\frac{1}{x}-\frac{1}{b}}\,dx.</math> |
| <math>r= R_{0}cos^{2}\theta</math> | |
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| <math>dr= -R_{0}2cos\theta sin\theta</math> | | We now make the substitution, |
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| | <math>x=b\cos^{2}\theta.\!</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>
| | We then obtain |
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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>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>I= R_{c}^{\frac{1}{2}}\int_{0}^{cos^{-1}\sqrt{\frac{R}{R_{c}}}} ( 1-{cos2\theta}) d\theta</math> | | <math>=2\sqrt{b}\int_{0}^{\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )}\sin^{2}\theta\,d\theta </math> |
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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>=\sqrt{b}\int_{0}^{\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )} ( 1-\cos{2\theta})\,d\theta</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> | | <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= 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> | | <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>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>=\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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| Let us consider <math>R_{c} \gg R</math> | | Let us consider the limit, <math>b\gg a.</math> We then have |
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| Then we have
| | <math>I\approx\frac{\pi}{2}\sqrt{b}-2\sqrt{a},</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> | | where we use the fact that <math>\cos^{-1}{x}\approx\frac{\pi}{2}-x.</math> |
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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>
| | Combining all of the above results, we get |
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| Setting, charge of <math>\alpha</math>particle = 2= <math>Z_{2}</math>(in general)
| | <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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| <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> | | 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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| 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>
| | <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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| Now putting <math>E= \frac{1}{2}mv^{2}</math>, veloctiy of the particle
| | 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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| <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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| Back to [[WKB Approximation#Problems|WKB Approximation]] | | Back to [[WKB Approximation#Problems|WKB Approximation]] |
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