Answer
$m = 2.6\times 10^{-25}~kg$
Work Step by Step
We can find an expression for the speed $v$:
$\frac{mv^2}{r} = qvB$
$v = \frac{q~B~r}{m}$
The particle has a kinetic energy of $q~\Delta V$. We can find the mass $m$:
$K = q~\Delta V$
$\frac{1}{2}mv^2 = q~\Delta V$
$\frac{1}{2}(m)(\frac{q~B~r}{m})^2 = q~\Delta V$
$m = \frac{q~B^2~r^2}{2~\Delta V}$
$m = \frac{(1.6\times 10^{-19}~C)~(1.2~T)^2~(0.125~m)^2}{(2)~(7000~V)}$
$m = 2.6\times 10^{-25}~kg$
