Phy5645/Gamowfactor

From the WKB apporximation we know that at the turning point, $E=V(x)=V_{coul}={\frac {1}{4\pi \epsilon _{0}}}{\frac {2z_{1}e^{2}}{R_{c}}}$

$R_{c}={\frac {1}{4\pi \epsilon _{0}}}{\frac {2z_{1}e^{2}}{E}}$

Now the Transition probabilty $T\cong \Theta ^{2}$, where $\Theta =e^{-\int _{b}^{a}q(x)dx}$

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

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

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.

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