Phy5645/Gamowfactor: Difference between revisions

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where <math>p(x)=\frac{1}{\hbar}\sqrt{2m\left(V(x)-E\right)}.</math>
where <math>p(x)=\frac{1}{\hbar}\sqrt{2m\left(V(x)-E\right)}.</math>


<math>\Theta ^{2} = e^{-2\int_{b}^{a} q(x)dx}</math>
We now evaluate the integral appearing in the exponential.
<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>


In the present problem <math>b= R</math> and <math>a = R_{c}</math>  
<math>=\sqrt{\frac{4mz_{1}e^{2}}{\hbar^2}}\int_{a}^{b} \sqrt{\frac{1}{x}-\frac{1}{b}}\,dx</math>


Now,
Let us define
<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>= \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>
<math>I=\int_{a}^{b} \sqrt{\frac{1}{x}-\frac{1}{b}}\,dx.</math>


let, <math>I = \int_{R}^{R_{c}} \left [ \frac{1}{r} - \frac{1}{R_{c}}\right ]^{\frac{1}{2}}dr</math>
We now make the substitution,
<math>x=b\cos^{2}\theta.\!</math>  


We then obtain


Put,
<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>
<math>r= R_{0}cos^{2}\theta</math>  
and


<math>dr= -R_{0}2cos\theta sin\theta</math>
<math>=2\sqrt{b}\int_{0}^{\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )}\sin^{2}\theta\,d\theta </math>


<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}\int_{0}^{\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )} ( 1-\cos{2\theta})\,d\theta</math>


<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 [ \theta - \sin\theta \cos\theta  \right ]_{0}^{\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )}</math>


<math>I= R_{c}^{\frac{1}{2}}\int_{0}^{cos^{-1}\sqrt{\frac{R}{R_{c}}}} ( 1-{cos2\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>


<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}\left [\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )-\sqrt{\frac{a}{b}}\sqrt{1-\frac{a}{b}} \right ]</math>


<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>
Let us consider the limit, <math>b\gg a.</math> We then have


<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>I\approx\frac{\pi}{2}\sqrt{b}-2\sqrt{a},</math>


<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>
where we use the fact that <math>\cos^{-1}{x}\approx\frac{\pi}{2}-x.</math>


Let us consider <math>R_{c} \gg R</math>
Combining all of the above results, we get


Then we have
<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>


<math>I\cong \sqrt{R_{c}}\left(cos^{-1}\sqrt{\frac{R}{R_{c}}}-\sqrt{\frac{R}{R_{c}}} \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


where <math>cos^{-1}\sqrt{\frac{R}{R_{c}}} \cong \frac{\pi}{2} - \left(\frac{R}{R_{c}}\right)^{\frac{1}{2}}</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>


Setting, charge of <math>\alpha</math>particle = 2= <math>Z_{2}</math>(in general)
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.
 
<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>
 
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>
 
Now putting <math>E= \frac{1}{2}mv^{2}</math>, veloctiy of the particle
 
<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>
 
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.


Back to [[WKB Approximation#Problems|WKB Approximation]]
Back to [[WKB Approximation#Problems|WKB Approximation]]

Latest revision as of 23:29, 14 January 2014

At the turning point,

so that

Within the WKB approximation, the transmission probability is given by

where

We now evaluate the integral appearing in the exponential.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sqrt{\frac{4mz_{1}e^{2}}{\hbar^2}}\int_{a}^{b} \sqrt{\frac{1}{x}-\frac{1}{b}}\,dx}

Let us define

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I=\int_{a}^{b} \sqrt{\frac{1}{x}-\frac{1}{b}}\,dx.}

We now make the substitution,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=b\cos^{2}\theta.\!}

We then obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =2\sqrt{b}\int_{0}^{\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )}\sin^{2}\theta\,d\theta }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sqrt{b}\int_{0}^{\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )} ( 1-\cos{2\theta})\,d\theta}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sqrt{b}\left [ \theta - \sin\theta \cos\theta \right ]_{0}^{\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 \}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sqrt{b}\left [\cos^{-1}\left (\sqrt{\frac{a}{b}}\right )-\sqrt{\frac{a}{b}}\sqrt{1-\frac{a}{b}} \right ]}

Let us consider the limit, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b\gg a.} We then have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I\approx\frac{\pi}{2}\sqrt{b}-2\sqrt{a},}

where we use the fact that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos^{-1}{x}\approx\frac{\pi}{2}-x.}

Combining all of the above results, we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 ).}

We may express this in terms of the velocity of the alpha particle by noting that the kinetic energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E=\tfrac{1}{2}mv^{2}.} Doing so, we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T=\exp\left (-\frac{4\pi z_{1}e^{2}}{\hbar v} \right )\exp\left (\frac{8e}{\hbar}\sqrt{z_{1}ma}\right ).}

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

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