Answer
a) $4.79\;\rm cm$
b) $5.4\times10^{-7}\;\rm s$
Work Step by Step
a)
We know that the doubly charged helium atom will move perpendicularly to a $0.240\;\rm T$-magnetic field which means that its radius of curvature is given by
$$F_B=ma=\dfrac{mv^2}{r}$$
We know that the magnetic force is given by
$$F_B=qBv$$
Thus,
$$qB v =\dfrac{mv^2}{r}$$
Thus,
$$r=\dfrac{mv}{qB}\tag 1$$
Now we need to find $v$.
We know that the ion is accelerated from rest by a potential difference which means that its potential energy, after enough time, is completely converted to kinetic energy.
Thus,
$$PE_i=KE_f$$
$$qV=\frac{1}{2}mv^2$$
Solving for $v$;
$$v=\sqrt{\dfrac{2qV}{m}}\tag 2$$
Plugging into (1);
$$r=\dfrac{m }{qB}\sqrt{\dfrac{2qV}{m}}=\sqrt{\dfrac{2qVm^2}{q^2B^2m}}$$
$$r=\sqrt{\dfrac{2 Vm }{q B^2 }}\tag 3$$
Plugging the known;
$$r=\sqrt{\dfrac{2 \cdot 3200\cdot 6.6\times10^{-27}}{2\cdot 1.6\times10^{-19} \cdot 0.240^2 }}$$
$$r=0.0479\;\rm m\approx\color{red}{\bf 4.79}\;\rm cm$$
b)
The period of revolution is given by the time it takes to make one complete circle.
Thus,
$$T=\dfrac{2\pi r}{v}$$
Plugging from (2) and (3);
$$T=\dfrac{2\pi \sqrt{\dfrac{2 Vm }{q B^2 }}}{\sqrt{\dfrac{2qV}{m}}}=2\pi\sqrt{\dfrac{2 Vm }{q B^2 } \cdot \dfrac{m}{2qV}}=\dfrac{2\pi m}{qB}$$
Plugging the known;
$$T=\dfrac{2\pi\cdot 6.6\times10^{-27}}{2\cdot 1.6\times10^{-19} \cdot 0.240}$$
$$T=\color{red}{\bf 5.4\times10^{-7}}\;\rm s$$
