Question

A singly charged ion 7 Li (an isotope of lithium) has a mass of 1.16 x 10 –²⁶ kg. It is accelerated through a potential difference of 220 V and then enters a magnetic field with magnitude 0.723 T perpendicular to the path of the ion. What is the radius of the ion's path in the magnetic field?

Answer 1

$$r=\frac{mv}{qB}=\frac{m\sqrt{2qV/m}}{qB}=\sqrt{\frac{2mV}{qB^2}}$$

$$r=\sqrt{\frac{2*1.16*10^{-26}*220}{1.6*10^{-19}*0.723^2}}=0.0078\:\rm m=7.8\; mm$$