Answer
$m = 4.00~u$
The particle is an alpha particle, which is the nucleus of a helium atom.
Work Step by Step
We can find $\gamma$:
$\gamma = \frac{1}{\sqrt{1-\beta^2}}$
$\gamma = \frac{1}{\sqrt{1-0.710^2}}$
$\gamma = 1.42$
We can find the mass $m$:
$r = \frac{\gamma ~m~v}{\vert q \vert B}$
$m = \frac{r~\vert q \vert B}{\gamma~v}$
$m = \frac{(6.28~m)~(2)(1.6\times 10^{-19}~C) (1.00~T)}{(1.42)(0.710)(3.0\times 10^8~m/s)}$
$m = 6.644\times 10^{-27}~kg$
$m = (6.644\times 10^{-27}~kg)(\frac{1~u}{1.66\times 10^{-27}~kg})$
$m = 4.00~u$
We can assume that this particle has two protons and two neutrons.
Therefore, the particle is an alpha particle, which is the nucleus of a helium atom.
