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 300 V and then enters a magnetic field with magnitude 0.742 T perpendicular to the path of the ion. What is the radius of the ion's path in the magnetic field?

8.87mm

Step-by-step explanation:

If the singled charged ion accelerates through a magnetic field, the force acting on the ion is expressed as;

F = qvBsin (theta)

Since the magnetic field is perpendicular to the path of the iron, angle that the ion made with the field will be 90°

F = qvBsin90°

F = qvB ... (1)

q is the charge on the ion

v is the velocity possessed by the charge

B is the magnetic Field

Also according to Newton's second law, the force experienced by the ion can also be expressed as;

F = ma = mv²/R ... (2)

Where a is the centripetal acceleration = v²/R

m is the mass of the ion

R is the radius

Equating both value of the force to get the radius;

qvB = mv²/R

qB = mv/R

R = mv/qB ... (3)

Given m = 1.16*10^-26kg

q = 1.6*10^-19C

B = 0.742T

v = ?

Before we can get the radius R, we need to get the velocity of the charge in the wire using the relationship.

1/2mv² = eV (since the ion possess kinetic energy and potential difference V)

From the equation, v² = 2eV/m

v = √2eV/m

v = √2*1.6*10^-19*300/1.16*10^-26

v = √8.28*10^9

v = 90971.77m/s

Substituting v = 90971.77m/s into equation 3 to get the radius R of the ion's path in the field will give;

R = mv/qB

R = 1.16*10^-26 * 90971.77/1.6*10^-19*0.742

R = 1.056*10^-21/1.19*10^-19

R = 0.00887m

R = 8.87mm