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Chapter 4: Q31E (page 714)

Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 155 V/m and the magnetic field is 0.0315 T. The ions next enter a uniform magnetic field of magnitude 0.0175 T that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) If the radius of the path of the ions in the second magnetic field is 17.5 cm, what is their mass?

Short Answer

Expert verified

a) They are moving 4920.6349m / s when they emerge in the velocity sector

b) Their mass is 9.9581脳10-26kg

Step by step solution

01

Determine the velocity and the arc path

The velocity of the particle is

v=EB

And the arc trajectory of the particle is

R=mvqB

02

Determine the velocity of the particle

(a)

The velocity of the particle in that region is

v=EB=1550.0315=4920.6349m/s

Therefore, the velocity of the particle is 4920.6349 m / s

03

Determine the mass of the particle

(b)

The arc path of the particle is calculated by

R=mvqB

Therefore, put the value we get

R=mvqB=(0.175)(1.6脳10鈭�19)(0.0175)4920.6349=9.9581脳10鈭�26kg

Therefore, the mass of the particle is 9.9581脳10-26kg

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