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

$\Theta ^{2}=e^{-2\int _{b}^{a}q(x)dx}$

In the present problem $b=R$ and $a=R_{c}$

Now, $\int _{R}^{R_{c}}\left({\frac {2m}{\hbar ^{2}}}\right)^{\frac {1}{2}}(V(x)-E)^{\frac {1}{2}}dr=\left({\frac {2m}{\hbar ^{2}}}\right)^{\frac {1}{2}}\int _{R}^{R_{c}}\left({\frac {1}{4\pi \epsilon _{0}}}{\frac {2z_{1}e^{2}}{r}}-E\right)^{\frac {1}{2}}dr$

$=\left({\frac {2m}{\hbar ^{2}}}\right)^{\frac {1}{2}}\left({\frac {2z_{1}e^{2}}{4\pi \epsilon _{0}}}\right)^{\frac {1}{2}}\int _{R}^{R_{c}}\left[{\frac {1}{r}}-{\frac {1}{R_{c}}}\right]^{\frac {1}{2}}dr$

let, $I=\int _{R}^{R_{c}}\left[{\frac {1}{r}}-{\frac {1}{R_{c}}}\right]^{\frac {1}{2}}dr$

Put, $r=R_{0}cos^{2}\theta$ and

$dr=-R_{0}2cos\theta sin\theta$

$I=2\int _{0}^{cos^{-1}{\sqrt {\frac {R}{R_{c}}}}}\left({\frac {R_{c}sin^{2}\theta }{cos^{2}\theta }}\right)^{\frac {1}{2}}cos\theta sin\theta d\theta$

$2R_{c}^{\frac {1}{2}}\int _{0}^{cos^{-1}{\sqrt {\frac {R}{R_{c}}}}}sin^{2}\theta d\theta$

$I=R_{c}^{\frac {1}{2}}\int _{0}^{cos^{-1}{\sqrt {\frac {R}{R_{c}}}}}(1-{cos2\theta })d\theta$

$I=R_{c}^{\frac {1}{2}}\left[\theta -sin\theta cos\theta \right]_{0}^{cos^{-1}{\sqrt {\frac {R}{R_{c}}}}}$

$I=R_{c}^{\frac {1}{2}}\left[cos^{-1}{\sqrt {\frac {R}{R_{c}}}}-sin\left(cos^{-1}{\sqrt {\frac {R}{R_{c}}}}\right)cos\left(cos^{-1}{\sqrt {\frac {R}{R_{c}}}}\right)\right]$

$I=R_{c}^{\frac {1}{2}}\left[cos^{-1}{\sqrt {\frac {R}{R_{c}}}}-{\sqrt {\frac {R}{R_{c}}}}{\sqrt {1-{\frac {R}{R_{c}}}}}\right]$

$I=R_{c}^{\frac {1}{2}}\left[cos^{-1}{\sqrt {\frac {R}{R_{c}}}}-{\sqrt {{\frac {R}{R_{c}}}-\left({\frac {R}{R_{c}}}\right)^{2}}}\right]$

Let us consider $R_{c}\gg R$

Then we have

$I\cong {\sqrt {R_{c}}}\left(cos^{-1}{\sqrt {\frac {R}{R_{c}}}}-{\sqrt {\frac {R}{R_{c}}}}\right)$

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

Setting, charge of $\alpha$particle = 2= $Z_{2}$(in general)

$\int q(x)dx=\left({\frac {2Mz_{1}z_{2}e^{2}R_{c}}{\hbar ^{2}4\pi \epsilon _{0}}}\right)^{\frac {1}{2}}\left[{\frac {\pi }{2}}-2\left({\frac {R}{R_{c}}}\right)^{\frac {1}{2}}\right]$

Now $T\cong e^{-2\int q(x)dx}=exp\left[-{\frac {\pi z_{1}z_{2}e^{2}}{\hbar 4\pi \epsilon _{0}}}\left({\frac {2M}{e}}\right)^{2}+{\frac {4}{\hbar }}\left({\frac {2z_{1}z_{2}e^{2}MR}{4\pi \epsilon _{0}}}\right)^{\frac {1}{2}}\right]$

Now putting $E={\frac {1}{2}}mv^{2}$, veloctiy of the particle

$T\cong exp\left({\frac {-2\pi z_{1}z_{2}e^{2}}{4\pi \epsilon _{0}\hbar v}}\right)exp\left({\frac {32z_{1}z_{2}e^{2}MR}{4\pi \epsilon _{0}\hbar ^{2}}}\right)^{\frac {1}{2}}$

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