Phy5645/Gamowfactor: Difference between revisions

Latest revision as of 23:29, 14 January 2014

At the turning point,

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

so that

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

Within the WKB approximation, the transmission probability is given by

$T=\exp \left[-2\int _{a}^{b}p(x)\,dx\right],$

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

We now evaluate the integral appearing in the exponential.

$\int _{a}^{b}p(x)\,dx={\sqrt {\frac {2m}{\hbar ^{2}}}}\int _{a}^{b}{\sqrt {V(x)-E}}\,dx={\sqrt {\frac {2m}{\hbar ^{2}}}}\int _{a}^{b}{\sqrt {{\frac {2z_{1}e^{2}}{x}}-E}}\,dx$

$={\sqrt {\frac {4mz_{1}e^{2}}{\hbar ^{2}}}}\int _{a}^{b}{\sqrt {{\frac {1}{x}}-{\frac {1}{b}}}}\,dx$

Let us define

$I=\int _{a}^{b}{\sqrt {{\frac {1}{x}}-{\frac {1}{b}}}}\,dx.$

We now make the substitution,

$x=b\cos ^{2}\theta .\!$

We then obtain

$I=2\int _{0}^{\cos ^{-1}\left({\sqrt {\frac {a}{b}}}\right)}{\sqrt {\frac {b\sin ^{2}\theta }{\cos ^{2}\theta }}}\cos \theta \sin \theta \,d\theta$

$=2{\sqrt {b}}\int _{0}^{\cos ^{-1}\left({\sqrt {\frac {a}{b}}}\right)}\sin ^{2}\theta \,d\theta$

$={\sqrt {b}}\int _{0}^{\cos ^{-1}\left({\sqrt {\frac {a}{b}}}\right)}(1-\cos {2\theta })\,d\theta$

$={\sqrt {b}}\left[\theta -\sin \theta \cos \theta \right]_{0}^{\cos ^{-1}\left({\sqrt {\frac {a}{b}}}\right)}$

$={\sqrt {b}}\left\{\cos ^{-1}\left({\sqrt {\frac {a}{b}}}\right)-\sin \left[\cos ^{-1}\left({\sqrt {\frac {a}{b}}}\right)\right]\cos \left[\cos ^{-1}\left({\sqrt {\frac {a}{b}}}\right)\right]\right\}$

$={\sqrt {b}}\left[\cos ^{-1}\left({\sqrt {\frac {a}{b}}}\right)-{\sqrt {\frac {a}{b}}}{\sqrt {1-{\frac {a}{b}}}}\right]$

Let us consider the limit, $b\gg a.$ We then have

$I\approx {\frac {\pi }{2}}{\sqrt {b}}-2{\sqrt {a}},$

where we use the fact that $\cos ^{-1}{x}\approx {\frac {\pi }{2}}-x.$

Combining all of the above results, we get

$T=\exp \left(-{\frac {2\pi z_{1}e^{2}}{\hbar }}{\sqrt {\frac {2m}{E}}}+{\frac {4}{\hbar }}{\sqrt {4mz_{1}e^{2}a}}\right).$

We may express this in terms of the velocity of the alpha particle by noting that the kinetic energy $E={\tfrac {1}{2}}mv^{2}.$ Doing so, we obtain

$T=\exp \left(-{\frac {4\pi z_{1}e^{2}}{\hbar v}}\right)\exp \left({\frac {8e}{\hbar }}{\sqrt {z_{1}ma}}\right).$

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

@@ Line 13: / Line 13: @@
 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]]

Phy5645/Gamowfactor: Difference between revisions

Latest revision as of 23:29, 14 January 2014

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