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 15 fm in diameter. What is the speed of the alpha particle when it is detected in the laboratory? Assume the thorium nucleus remains at rest.

Answer 1

Answer:

Velocity will be $v=4.06\times10^7m/sec$

Explanation:

We have given mass of alpha particle m = 4 u $=4\times 1.67\times 10^{-27}kg=6.68\times 10^{-27}kg$

Charge on alpha particle $q=2e=2\times 1.6\times 10^{-19}C=3.2\times 10^{-19}C$

Charge on thorium particle $=90\times 1.6\times 10^{-19}=144\times 10^{-19}C$

Diameter is given as d = 15 fm

So radius $r=\frac{d}{2}=\frac{15}{2}=7.5fm=7.5\times 10^{-15}m$

Potential energy is given by $E=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r}=\frac{Kq_1q_2}{r}=\frac{9\times 10^9\times 144\times 10^{-19}\times 3.2\times 10^{-19}}{7.5\times 10^{-15}}=5.529\times 10^{-12}J$

From energy conservation $\frac{1}{2}mv^2=5.529\times 10^{-12}$

$\frac{1}{2}\times 6.68\times 10^{-27}v^2=5.529\times 10^{-12}$

$v=4.06\times10^7m/sec$