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 250 V and then enters a magnetic field with magnitude 0.878 T perpendicular to the path of the ion. What is the radius of the ion's path in the magnetic field?

Answer 1

Answer:

Radius = 6.877mm

Step-by-step explanation:

From the question, we have a single charged ion of 7Li with;

Mass(m) = 1.16 × 10^(−26) kg

Potential difference(V) = 250 V

Electron charge of e = 1.6 x 10^(-19)

To solve this question, we'll equate the kinetic energy to potential energy. Thus;

(1/2)(mv²) = eV

Thus;

Velocity (v) =√(2eV/m)

v = √((2 x 1.6 x 10^(-19) x 250)/1.16 × 10^(−26))

v = √(689.655 x 10^(7) = 8.3 x 10⁴ m/s

Since it enters a magnetic field with magnitude B = 0.874 T perpendicular to the path of the ion, thus θ = 90°

Now,we know that;

F = qvBsinθ

Thus F = qvBsin(90)

Sin 90° = 1 ; thus, F =qvB

Also, from Newton's second law, F=ma, also, we know that radial acceleration a = v²/r

So F = mv²/r

Thus equating the 2 forces, we have ;

qvB = mv²/r

So making r the subject, we have;

r = mv/qB = (1.16 × 10^(−26) x 8.3 x 10⁴)/(1.6 x 10^(-19) x 0.874)

= 6.877 x 10^(-3)m = 6.877mm