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• He⁺ ions with a charge of [tex]+e[/tex] move at [tex]1.00 \times 10^5 \, \text{m/s}[/tex] in a straight line through a velocity selector. Suppose the He⁺ ions are replaced with He²⁺ ions in which both electrons have been removed from the helium atom and the ion charge is [tex]+2e[/tex]. At what speed must the He²⁺ ions travel through the same velocity selector in order to move in a straight line?

Sagot :

GinnyAnswer
To solve this problem, we need to understand how a velocity selector works. A velocity selector typically has both electric and magnetic fields. The electric field exerts a force on the charged particles in the direction of the field, while the magnetic field exerts a force perpendicular to both the field and the particle's velocity.

For a particle to move in a straight line through the velocity selector, the electric force and magnetic force must balance each other. The electric force [tex]\((F_e)\)[/tex] on a charged particle is given by:
[tex]\[ F_e = qE \][/tex]
where:
- [tex]\( q \)[/tex] is the charge of the particle,
- [tex]\( E \)[/tex] is the electric field strength.

The magnetic force [tex]\((F_m)\)[/tex] on a charged particle moving with velocity [tex]\((v)\)[/tex] perpendicular to a magnetic field [tex]\((B)\)[/tex] is given by:
[tex]\[ F_m = qvB \][/tex]

For the particle to move in a straight line, these forces must be equal:
[tex]\[ qE = qvB \][/tex]
We can simplify this to:
[tex]\[ E = vB \][/tex]

Given data:
- The speed of He+ ions ([tex]\( v_{He+} \)[/tex]) is [tex]\( 1.00 \times 10^5 \, \text{m/s} \)[/tex].
- The charge of He+ ions ([tex]\( q_{He+} \)[/tex]) is [tex]\( +e \)[/tex].
- The charge of He2+ ions ([tex]\( q_{He2+} \)[/tex]) is [tex]\( +2e \)[/tex].

We need to find the speed of He2+ ions ([tex]\( v_{He2+} \)[/tex]) so that they also move in a straight line through the velocity selector.

For He+ ions:
[tex]\[ E = v_{He+} B \][/tex]
Given that [tex]\( v_{He+} = 1.00 \times 10^5 \, \text{m/s} \)[/tex], we have:
[tex]\[ E = (1.00 \times 10^5)B \][/tex]

For He2+ ions, to balance the forces and maintain a straight path:
[tex]\[ E = v_{He2+} B \][/tex]

Since the electric field [tex]\( E \)[/tex] in the velocity selector remains the same for both ions, we equate the two expressions for [tex]\( E \)[/tex]:
[tex]\[ (1.00 \times 10^5) B = v_{He2+} B \][/tex]

We can cancel [tex]\( B \)[/tex] from both sides:
[tex]\[ 1.00 \times 10^5 = v_{He2+} \][/tex]

So to maintain the straight path and balance the magnetic and electric forces for He2+ ions, they must travel at:
[tex]\[ v_{He2+} = \frac{1.00 \times 10^5}{2} \][/tex]

Therefore, the speed at which the He2+ ions must travel through the velocity selector is:
[tex]\[ v_{He2+} = 5.00 \times 10^4 \, \text{m/s} \][/tex]
or
[tex]\[ v_{He2+} = 50000.0 \, \text{m/s} \][/tex]
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