Finding velocity of doubly charged He atom from accelerating voltage

[MadMustang129]

Homework Statement

A doubly charged helium atom whose mass is 6.6 \times 10^{ - 27} {\rm{kg}} is accelerated by a voltage of 2800 V.
What will be its radius of curvature if it moves in a plane perpendicular to a uniform 0.370 -T field?
What is its period of revolution?

Homework Equations

F=qvB
centripetal acceleration = (v^2)/r
P= 1/f

The Attempt at a Solution

I know that F=ma, and a= (v^2)/r, and so r will = mv/qB.

My principle question is how can I find out the particle's initial velocity in the magnetic field based on the voltage that accelerates the atom? Also, what is the charge of a "doubly charged" He atom and how do you know? I understand the magnetism principles behind it but am having difficulty figuring out the speed at which the particle enters the mass spectrometer. Thank you... I appreciate the help!

 

[alphysicist]

Hi MadMustang129,

Homework Statement

A doubly charged helium atom whose mass is 6.6 \times 10^{ - 27} {\rm{kg}} is accelerated by a voltage of 2800 V.
What will be its radius of curvature if it moves in a plane perpendicular to a uniform 0.370 -T field?
What is its period of revolution?

Homework Equations

F=qvB
centripetal acceleration = (v^2)/r
P= 1/f

The Attempt at a Solution

I know that F=ma, and a= (v^2)/r, and so r will = mv/qB.

My principle question is how can I find out the particle's initial velocity in the magnetic field based on the voltage that accelerates the atom?

Try applying conservation of energy to the motion before it reaches the magnetic field. What speed does that give?

Also, what is the charge of a "doubly charged" He atom and how do you know?

Doubly charged means that the magnitude of its charge is (+2 e), so that it is twice the magnitude of an electron.