Phy5645/Gamowfactor

Calculation of Gamow Factor for Alpha decay of Nuclei

Since the $\alpha$-decay happens in the nulcie then we can assume that an $\alpha$-decay is formed in the nucleus just before its emission (although $\alpha$ 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. Outside the range of the nuclear interaction would be coulombic. So the coulomb interaction is,

$V_{coul}={\frac {1}{4\pi \epsilon _{0}}}{\frac {2z_{1}e^{2}}{r}}$

where $Z_{1}$ is the atomic number of the rest of the nucleus(after decay).

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.