In the form of radioactive decay known as alpha decay, an unstable nucleus emits a helium-atom nucleus, which is called an alpha particle. An alpha particle contains two protons and two neutrons, thus having mass$m=4u$and charge $q=2e$. Suppose a uranium nucleus with $92$protons decays into thorium, with $90$protons, and an alpha particle. The alpha particle is initially at rest at the surface of the thorium nucleus, which is $15fm$in diameter. What is the speed of the alpha particle when it is detected in the laboratory? Assume the thorium nucleus remains at rest

Given :

An alpha particle contains mass : $m=4u$and charge $q=2e$

Uranium nucleus with protons :$92$

Protons with Thorium : $90$

Diameter of Thorium nucleus : $15fm$

Theory used :

From the equation of conservation of energy, we have :$\frac{1}{2}m{v}^2=q∆V$

Because the detection in the laboratory takes place so distant from the nucleus, we'll assume the potential at the moment of detection is zero.

As a result, the potential difference is equal to the potential at the nucleus' surface. To get the latter, we'll utilize the traditional approach of calculating the potential as

$V=\frac{2·92ke}{d}$

where$d$is the nucleus' diameter. We're ignoring the fact that the alpha particle has a nonzero radius; nevertheless, because we don't know what it is, and because of its much lower size, we may fairly assume it is zero.

Rewriting the conservation of energy law, we get speed :

$v=\sqrt{\frac{2q∆V}{m}}$

where the charge of the alpha particle is $q$.

Substitute for $∆V\mathrm{and}q$, and we'll get :

$v=\sqrt{\frac{2\times 2e\times 2\times 92ke}{md}}$

This speed can be calculated numerically to be :

localid="1650514833125" $$\begin{gathered} v=\sqrt{\frac{736\times {(1.6\times {10}^{-19})}^2\times 9\times {10}^9}{2\times 1.66\times {10}^{-27}\times 1.5\times {10}^{-14}}} \\ =\boxed{\mathbf{5}\mathbf{.}\mathbf{84}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}} \end{gathered}$$