Question

A singly charged ion of 7Li (an isotope of lithium) has a mass of 1.16×10−26 kg . It is accelerated through a potential difference of 290 V and then enters a magnetic field with magnitude 0.710 T perpendicular to the path of the ion. What is the radius of the ion's path in the magnetic field?

Answer 1

0.00645 m

Step-by-step explanation:

m = Mass of lithium =
`1.16* 10^(-26)\ kg`

V = Voltage = 290 V

r = Radius of path

B = Magnetic field = 0.710 T

q = Charge of electron =
`1.6* 10^(-19)\ C`

v = Velocity of mass

The kinetic energy in moving a charge and the work done by the charge will balance each other

`(1)/(2)mv^2=qV\\\Rightarrow v=\sqrt{(2qV)/(m)}\\\Rightarrow v=\sqrt{(1* 1.6* 10^(-19)* 290)/(1.16* 10^(-26))}\\\Rightarrow v=63245.5532\ m/s`

The velocity by which the charge is moving is 63245.5532 m/s

The force on moving a charge and the centripetal force will balance each other

`F=F_c\\\Rightarrow Bqv=m(v^2)/(r)\\\Rightarrow r=(mv)/(Bq)\\\Rightarrow r=(1.16* 10^(-26)* 63245.5532)/(0.71* 1.6* 10^(-19))\\\Rightarrow r=0.00645\ m`

The radius of the magnetic field is 0.00645 m