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?

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