Phy5645/Gamowfactor

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.

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