Question

A ²²²Rn atom (radon) in a 0.75 T magnetic field undergoes radioactive decay, emitting an alpha particle in a direction perpendicular to $\overrightarrow{B}$. The alpha particle begins cyclotron motion with a radius of 45 cm. With what energy, in MeV, was the alpha particle emitted?

Short answer

The energy of alpha particle emitted is 5.49 MeV

Step-by-step solution

Step 1. Given information is :Magnetic field = 0.74 TRadius of initial cyclotron motion = 45 cm

We need to find out the energy of alpha particle emitted.

Step 2. Using the equilibrium condition between centripetal force and magnetic force

$$\begin{gathered} F_{cp}=F_B \\ \frac{m{v}^2}{r}=qvB \\ v=\frac{qBr}{m} \\ {E}_k=\frac{m{v}^2}{2} \\ {E}_k=\frac{m{q}^2{B}^2{r}^2}{2m^2} \\ {E}_k=\frac{q^2{B}^2{r}^2}{2m} \\ {E}_k=\frac{{(2\times 1.6\times {10}^{-19})}^2(0.75{)}^2{(0.45)}^2}{2\times 6.64\times {10}^{-27}}J \\ {E}_k=8.78\times {10}^{-13}J=5.49MeV \end{gathered}$$