Answer
${\bf 5.46}\;\rm MeV$
Work Step by Step
To find the energy of the alpha particle emitted during the radioactive decay of $^{222}\rm{Rn}$, we can use the principles of cyclotron motion in a magnetic field.
The centripetal force acting on the alpha particle in its cyclotron motion is provided by the magnetic force. The formula for the magnetic force on a charged particle moving perpendicular to a magnetic field is:
$$F = qvB $$
and according to Newton's second law,
$$F = \frac{mv^2}{r} $$
Thus,
$$qvB = \frac{mv^2}{r}$$
We can solve for the velocity of the alpha particle:
$$ v = \frac{qBr}{m} \tag 1$$
The kinetic energy of the alpha particle is given by:
$$K = \frac{1}{2}mv^2 $$
Plug from (1);
$$K = \frac{1}{2}m\left[\frac{qBr}{m} \right]^2 $$
The charge of an alpha particle is twice the charge of a proton, $q=2e$, and its mass is 4 times the mass of one proton, $m=4m_p$
$$K = \frac{4e^2B^2r^2}{2(4m_p)} $$
$$K = \frac{ e^2B^2r^2}{ 2m_p} $$
Plug the known;
$$K = \frac{ (1.6\times 10^{-19})^2(0.75)^2(0.45)^2}{ 2 (1.67\times 10^{-27})} $$
$$K=\bf 8.73054\times 10^{-13}\;\rm J=\color{red}{\bf 5.46}\;\rm MeV$$
