Answer
$6.7\times 10^{6}\ m/s$
Work Step by Step
Given that:
Radial acceleration of an electron $a=3\times 10^{14}\ m/s^2$
Radius of circular path $r=15\ cm = 0.15\ m$
We know that centripetal acceleration $a=\frac{v^2}{r}$.
Rearranging and solving:
$v^2=ar$
$v=\sqrt {ar}$
$v=\sqrt{ (3\times 10^{14}\ m/s^2)(0.15\ m)}$
$v=6.7\times 10^{6}\ m/s$
