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Chapter 4: Q26E (page 913)

A singly charged ion of 7Li (an isotope of lithium) has a mass of 1.16 x 10-26 kg. It is accelerated through a potential difference of 220 V and then enters a magnetic field with magnitude 0.874 T perpendicular to the path of the ion. What is the radius of the ion’s path in the magnetic field?

Short Answer

Expert verified

The radius of the ion’s path in the magnetic field is 6.46mm

Step by step solution

01

Determine the kinetic energy and potential energy

The potential energy is equal to the kinetic energy

12mv2=eV

02

Determine the velocity of the particle

7Li has a mass is1.16 x 10-26 kg

Therefore, the value foris

v=2eVm=2(1.60×10−19C)(220V)1.16×10−26kg=7.79×104m/s

03

Determine the radius of the ion’s path in the magnetic field

The magnitude of the magnetic field is 0.874 T

Therefore, the force acting on the particle is

FB=qvB=ma

And the radial velocity is

role="math" localid="1668227699257" a=v2RqvB=mv2RThereforeR=mvqB

Therefore, by putting the value we get

R=6.46×10−3m=6.46mm

Therefore, the radius of the ion’s path in the magnetic field is 6.46mm

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