Question

In the radioactive decay of Eq. 21-13, a $U^{238}$ nucleus transforms

to$T{h}^{234}$and an ejected$H{e}^4$. (These are nuclei, not atoms, and thus

electrons are not involved.) When the separation between and

4He is$9.0\times {10}^{-15}\text{m}$, what are the magnitudes of (a) the electrostatic

force between them and (b) the acceleration of the particle?

Short answer

a) The electrostatic force between them is$F=5.1\times {10}^2\text{\hspace{0.17em}N}$

b) The acceleration of the particle is$a=7.7\times {10}^{28}\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Step-by-step solution

Write the given data

Consider the given chemical equation as:

$U^{238}\to T{h}^{234}+H{e}^4$

Separation between $T{h}^{234}$and$H{e}^4$ is$9\times {10}^{-15}\text{\hspace{0.17em}m}$

(a) Calculate the magnitudes of the electrostatic force between them

The nucleus of Helium has 2 protons and that of thorium has 90.

$\begin{array}{c}F=k\frac{q^2}{d^2}\\ =(9\times {10}^9\frac{{\text{Nm}}^{\text{2}}}{{\text{C}}^{\text{2}}})\frac{(2\times 1.60\times {10}^{-19}\text{\hspace{0.17em}C})(90\times 1.60\times {10}^{-19}\text{C})}{{(9.0\times {10}^{-15}\text{\hspace{0.17em}m})}^2}\\ =5.1\times {10}^2\text{\hspace{0.17em}N}\end{array}$

(b) Calculate the magnitudes of the acceleration of the particle

Estimating the helium nucleus mass as that of 4 protons (actually, that of 2

Protons and 2 neutrons, but the neutrons have approximately the same mass), Newton’s second law leads to

$\begin{array}{c}a=\frac{F}{m}\\ =\frac{5.1\times {10}^2\text{\hspace{0.17em}N}}{4(1.67\times {10}^{-27}\text{\hspace{0.17em}kg})}\\ =7.7\times {10}^{28}\frac{\text{m}}{{\text{s}}^{\text{2}}}\end{array}$