Answer
$K.E=1.7\times 10^7 eV$
Work Step by Step
$K.E=\frac{1}{2}mv^2$.............eq(1)
As centripetal force is produced to the magnetic force so these two forces are equal in magnitude i.e
$\frac{mv^2}{R}=qvB$
$mv=qBR$
Squaring both sides,we obtain
$(mv)^2=(qBR)^2$
or $mv^2=\frac{(qBR)^2}{m}$
Substituting this value in eq(1), we obtain:
$K.E=\frac{1}{2}\frac{(qBR)^2}{m}$
Now weplug in the known values to obtain:
$K.E=\frac{(1.602\times 10^{-19}\times1.2\times 0.5)^2}{2(1.67\times 10^{-27})}$
$K.E=2.76\times 10^{-12}J$
$K.E=\frac{2.76\times 10^{-12}}{1.602\times 10^{-19}}eV$
$K.E=1.7\times 10^7 eV$
