Question

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

Short answer

The speed of the alpha particle is$\mathbf{5}\mathbf{.}\mathbf{84}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$

Step-by-step solution

Given information and theory used 

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 :$\frac{1}{2}m{v}^2=q∆V$

Calculating the speed of the alpha particle 

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