Question

When radium-226 decays radioactively, it emits an alpha particle (the nucleus of helium), and the end product is radon-222. We can model this decay by thinking of the radium-226 as consisting of an alpha particle emitted from the surface of the spherically symmetric radon-222 nucleus, and we can treat the alpha particle as a point charge. The energy of the alpha particle has been measured in the laboratory and has been found to be$4.79MeV$ when the alpha particle is essentially infinitely far from the nucleus. Since radon is much heavier than the alpha particle, we can assume that there is no appreciable recoil of the radon after the decay. The radon nucleus contains 86 protons, while the alpha particle has 2 protons and the radium nucleus has protons. (a) What was the electric potential energy of the alpha–radon combination just before the decay, in MeV and in joules? (b) Use your result from part (a) to calculate the radius of the radon nucleus.

Short answer

a. The electrical potential energy of the alpha-radon combination just before decay is $\text{4.79 MeV}$and $7.66\times {10}^{-13}J$.

b. The radius of the radon nucleus is $5.17\times {10}^{-14}m$.

Step-by-step solution

(a) Determination of the electrical potential energy of alpha-radon combination just before decay.

The mechanical energy of the system is always conserved. The final energy of the system is the potential energy of the system just before the alpha particle is ejected.

$U=k\frac{qq\text{'}}{r}$

Energy conservation relation here is,

${\text{U}}_{\text{a}}{\text{+K}}_{\text{a}}{\text{=U}}_{\text{b}}{\text{+K}}_{\text{b}}\text{.}$

Thus, the potential energy is the final energy when the decay happens,

$\begin{array}{rcl}& & \text{U = 4.79 MeV}\\ & =& 7.66\times {10}^{-13}J\\ & & \end{array}$

(b) Comparison of the electric potential energy calculated in part (a) to the potential energy of the proton–electron pair in the hydrogen atom.

Now, the charge of the alpha particle is $=+2e$

And the charge of the radon nucleus is $=+86e$

Substitute all the values and solve for r,

role="math" localid="1655814415670" $\begin{array}{rcl}& & \text{r = k}\frac{\text{qq'}}{\text{U}}\\ & =& \frac{\left(8.99\times {10}^{9}N\times m^2/C^2\right)\left(2\right)\left(86\right){\left(1.60\times {10}^{-19}C\right)}^2}{7.66\times {10}^{-13}J}\\ & =& 5.17\times {10}^{-14}m\\ & & \end{array}$

Thus, the radius of the radon nucleus is$5.17\times {10}^{-14}m$.