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?

Expert verified

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

b) Their mass is $9.9581 \times 10^{- 26} kg$

Step by step solution

The velocity of the particle is

$v = \frac{E}{B}$

And the arc trajectory of the particle is

$R = \frac{m v}{q B}$

(a)

The velocity of the particle in that region is

$\begin{aligned} v = \frac{E}{B} \\ = \frac{155}{0.0315} \\ = 4920 .6349 m / s \end{aligned}$

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

(b)

The arc path of the particle is calculated by

$R = \frac{m v}{q B}$

Therefore, put the value we get

$\begin{aligned} R = \frac{mv}{qB} \\ = \frac{(0 .175) (1 .6 \times 10^{- 19}) (0 .0175)}{4920 .6349} \\ = 9 .9581 \times 10^{- 26} kg \end{aligned}$

Therefore, the mass of the particle is $9.9581 \times 10^{- 26} k g$

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