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'''Calculation of Gamow Factor for Alpha decay of Nuclei'''
At the turning point,


[[Image:Gamow.jpg|400px]]
<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).
Inside the nucleus the particle will experience nuclear interaction which mostly attractive and outside the nucleus the inetraction would be coulombic(replusive).


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,


<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>


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


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>  


<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>


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>


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>


<math>\Theta ^{2} = e^{-2\int_{b}^{a} q(x)dx}</math>
Let us define


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>


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>  


<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


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>


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


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>  
and


<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>


<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>


<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>


<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


<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>


<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>


<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


<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>


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


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>


<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.


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]]
 
Setting, charge of <math>\alpha</math>particle = 2= <math>Z_{2}</math>(in general)
 
<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.

Latest revision as of 23:29, 14 January 2014

At the turning point,

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=V(b)=\frac{2z_{1}e^{2}}{b},}

so 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 b=\frac{2z_{1}e^{2}}{E}.}

Within the WKB approximation, the transmission probability is given by

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 [-2\int_{a}^{b}p(x)\,dx\right ],}

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

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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