Question 21.69: In the radioactive decay of Eq. 21-13, a ^238U nucleus trans......
In the radioactive decay of Eq. 21-13, a ^{238}U nucleus transforms to ^{234}Th and an ejected ^4He. (These are nuclei, not atoms, and thus electrons are not involved.) When the separation between ^{234}Th and ^4He is 9.0 × 10^{-15} m, what are the magnitudes of (a) the electrostatic force between them and (b) the acceleration of the ^4He particle?
We are concerned with the charges in the nucleus (not the “orbiting” electrons, if there are any). The nucleus of Helium has 2 protons and that of thorium has 90.
(a) Equation 21-1 gives
\vec{F}=k \frac{q_1 q_2}{r^2} \hat{ r } (Coulomb’s law), (21-1)
F=k \frac{q^2}{r^2}=\frac{\left(8.99 \times 10^9 \,N \cdot m ^2 / C ^2\right)\left(2\left(1.60 \times 10^{-19} \,C \right)\right)\left(90\left(1.60 \times 10^{-19}\, C \right)\right)}{\left(9.0 \times 10^{-15} \,m \right)^2}=5.1 \times 10^2 \,N .
(b) 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
a=\frac{F}{m}=\frac{5.1 \times 10^2 \,N }{4c 1.67 \times 10^{-27} \,kgh }=7.7 \times 10^{28} \,m / s ^2 .