Asingly charged ion of 7li (an isotope of lithium which lost only one electron) has a mass of 1.16 ×10^-26 kg. it is accelerated through a potential difference of 224 v and then enters a magnetic field with magnitude 0.724 t perpendicular to the path of the ion. what is the radius of the ion’s path in the magnetic field? (give your answer in decimal using "mm"(millimeter) as unit)

Explanation:

It is given that,

Mass of lithium,

It is accelerated through a potential difference, V = 224 V

Uniform magnetic field, B = 0.724 T

Applying the conservation of energy as :

q is the charge on an electron

v = 78608.58 m/s

To find the radius of the ion's path in the magnetic field. The centripetal force is balanced by the magnetic force as :

r = 0.0078 meters

So, the radius of the path of the ion is 0.0078 meters. Hence, this is the required solution.

the closer the planet is to the sun, the larger the gravity