Answer
a) $6.0 * 10^5$ m/$s$
b) $3.6 * 10^{-2}$ m
c) $3.8 * 10^{-7} s$
Work Step by Step
a) $\Delta KE = qV$
$\frac{1}{2}mv^2 = qV $
$v= \sqrt \frac{2qV}{m}$
$v= \sqrt \frac{2*2*1.6*10^{-19} * 3700}{6.6*10^{-27}} $ = $6.0 * 10^5$ m/$s$
b) $F=qvB$
$\frac{mv^2}{r} = qvB$
$r = \frac{mv}{qB} = = \frac{6.6*10^{-27} * 6.0*10^5}{2*1.6*10^{-19}* .34} = 3.6 * 10^{-2}$ m
c) $T = \frac{2*pi*m}{qB} = \frac{2*3.14*6.6*10^{-27}}{2*1.6*10^{-19}*.34} = 3.8 * 10^{-7} s$
