Calculate Radius & Period of Revolution of Doubly Charged He Atom

[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.

 

[nlightner]

I would first clarify the information provided in the homework statement. The mass of the doubly charged helium atom given is 6.6 \times 10^{ - 27} {\rm{kg}}, which is the mass of a single helium atom. However, since it is doubly charged, the charge of the atom is +2e, where e is the elementary charge. This information is important in calculating the velocity of the atom.

To find the initial velocity of the atom, we can use the equation F=qvB, where F is the force exerted on the atom, q is its charge, v is its velocity, and B is the magnetic field. In this case, the force is provided by the voltage of 2800 V, so we can rearrange the equation to solve for v: v = F/(qB). Plugging in the values, we get v = (2800 V)/(2e)(0.370 T) = 1.5 \times 10^6 m/s.

Using this initial velocity, we can now calculate the radius of curvature using the equation r = mv/(qB). Plugging in the values, we get r = (6.6 \times 10^{ - 27} {\rm{kg}})(1.5 \times 10^6 m/s)/(2e)(0.370 T) = 2.8 \times 10^{-4} m.

To find the period of revolution, we can use the equation P = 1/f, where P is the period and f is the frequency. The frequency is given by the equation f = v/(2\pi r), where v is the velocity and r is the radius of curvature. Plugging in the values, we get f = (1.5 \times 10^6 m/s)/(2\pi)(2.8 \times 10^{-4} m) = 1.4 \times 10^9 Hz. Finally, the period is P = 1/(1.4 \times 10^9 Hz) = 7.1 \times 10^{-10} s.

In summary, the radius of curvature of the doubly charged helium atom in a uniform magnetic field of 0.370 T is 2.8 \times 10^{-4} m and its period of revolution is 7.1 \times 10^{-10} s.