Answer
The alpha particle must have an energy of $~~1.0~MeV$
Work Step by Step
Let the proton's mass be $m_0 = 1.0~u$
Let the proton's charge be $q_0 = 1.0~e$
Let the speed of the proton be $v_0$
We can write an expression for the radius of motion of the proton in a magnetic field:
$r_0 = \frac{m_0~v_0}{q_0 B} = v_0~\frac{m_0}{q_0 B}$
We can write an expression for the kinetic energy of the proton:
$K_0 = \frac{1}{2}m_0v_0^2$
We can write an expression for the radius of motion of the alpha particle in the magnetic field:
$r = \frac{4m_0~v}{2q_0 B} = 2v~\frac{m_0}{q_0 B}$
If $r = r_0$, then $v = \frac{v_0}{2}$
We can find the kinetic energy of the alpha particle:
$K = \frac{1}{2}(4m_0)(\frac{v_0}{2})^2$
$K = \frac{1}{2}m_0v_0^2$
$K=K_0$
$K = 1.0~MeV$
The alpha particle must have an energy of $~~1.0~MeV$.
