Explanation:
It is given that,
Mass of lithium, [tex]m=1.16\times 10^{-26}\ kg[/tex]
It is accelerated through a potential difference, V = 224 V
Uniform magnetic field, B = 0.724 T
Applying the conservation of energy as :
[tex]\dfrac{1}{2}mv^2=qV[/tex]
[tex]v=\sqrt{\dfrac{2qV}{m}}[/tex]
q is the charge on an electron
[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 224\ V}{1.16\times 10^{-26}\ kg}}[/tex]
v = 78608.58 m/s
[tex]v=7.86\times 10^4\ m/s[/tex]
To find the radius of the ion's path in the magnetic field. The centripetal force is balanced by the magnetic force as :
[tex]qvB=\dfrac{mv^2}{r}[/tex]
[tex]r=\dfrac{mv}{qB}[/tex]
[tex]r=\dfrac{1.16\times 10^{-26}\ kg\times 7.86\times 10^4\ m/s}{1.6\times 10^{-19}\ C\times 0.724\ T}[/tex]
r = 0.0078 meters
So, the radius of the path of the ion is 0.0078 meters. Hence, this is the required solution.
