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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 169 V/m and the magnetic field is 3.12×10⁻² T . The ions next enter a uniform magnetic field of magnitude 1.77×10⁻² 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?

Respuesta :

Answer:

A) 5416.67 m/s

B) 9.15 x 10^(-26) Kg

Explanation:

A) |E| = 169 V/m

|Bi| = 3.12×10⁻² T or 0.0312 T

|Bf| = 1.77×10⁻² T or 0.0177 T

Velocity is given by;

V = |E|/|Bi|

Thus; V = 169/0.0312 = 5416.67 m/s

B) Radius (R) is given by;

R = mV/qB

q is proton charge = 1.6 x 10^(-19)

R = 17.5cm = 0.175m

M is mass and unknown

B is Bf = 0.0177 T

Thus, to find the mass, let's make "m" the subject.

m = qBR/V = [1.6 x 10^(-19) x 0.0177 x 0.175]/5416.67 = 9.15 x 10^(-26) Kg

A) The ions moving when they emerge from the velocity

selector is 5416.67 m/s.

B) Their mass is [tex]9.15 x 10^(-26) Kg.[/tex]

Velocity

Part A)

The ions moving when they emerge from the velocity selector

is :

Given:

|E| = 169 V/m

|Bi| = 3.12×10⁻² T or 0.0312 T

|Bf| = 1.77×10⁻² T or 0.0177 T

Velocity :

V = |E|/|Bi|

V = 169/0.0312

V = 5416.67 m/s

Part B)

Radius (R) = mV/qB

Proton charge (q) = 1.6 x 10^(-19)

R = 17.5cm = 0.175m

M = mass and unknown

B is Bf = 0.0177 T

m = qBR/V

m = [1.6 x 10^(-19) x 0.0177 x 0.175]/5416.67

[tex]m = 9.15 x 10^(-26) Kg[/tex]

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