Question

A singly 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)

Answer 1

Explanation:

It is given that,

Mass of lithium, $m=1.16\times 10^{-26}\ kg$

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

Uniform magnetic field, B = 0.724 T

Applying the conservation of energy as :

$\dfrac{1}{2}mv^2=qV$

$v=\sqrt{\dfrac{2qV}{m}}$

q is the charge on an electron

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 224\ V}{1.16\times 10^{-26}\ kg}}$

v = 78608.58 m/s

$v=7.86\times 10^4\ m/s$

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

$qvB=\dfrac{mv^2}{r}$

$r=\dfrac{mv}{qB}$

$r=\dfrac{1.16\times 10^{-26}\ kg\times 7.86\times 10^4\ m/s}{1.6\times 10^{-19}\ C\times 0.724\ T}$

r = 0.0078 meters

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