Radius of curvature atom path

Substituting your value for v and the given values for m, q, and B gives approximately 2.01 x 10^-3 meters as the radius of curvature. In summary, the radius of curvature of the path of a doubly charged helium atom, accelerated through a potential difference of 4.00x 10^3 V, in a uniform 0.450 T magnetic field is approximately 2.01 x 10^-3 meters.
  • #1
superjen
26
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A doubly charged helium atom (mass = 6.68 x 10-27 kg) is accelerated through a potential difference of 4.00x 103 V. What will be the radius of curvature of the path of the atom if it is in a uniform 0.450 T magnetic field?


the equation i was using was
r = mV/|q|B

m = 6.68 x 10^-27
V = 4.00 x 10^3
B = 0.450T
q = 1.6 x 10^-19

I don't think this is right. any help pr tips?
Thanks :)
 
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  • #2
superjen said:
A doubly charged helium atom (mass = 6.68 x 10-27 kg) is accelerated through a potential difference of 4.00x 103 V. What will be the radius of curvature of the path of the atom if it is in a uniform 0.450 T magnetic field?

the equation i was using was
r = mV/|q|B

m = 6.68 x 10^-27
V = 4.00 x 10^3
B = 0.450T
q = 1.6 x 10^-19

I don't think this is right. any help pr tips?
Thanks :)

First figure the kinetic energy, from the work done on the charge.

W = q*ΔV

From ½mv² you can derive a value for v .

Then you can use your second equation derived from

F = mv² /R = qv*B

R = mv/qB
 
  • #3



Thank you for providing the equation you were using. Your calculation is actually correct! Let me explain how I arrived at the same answer as you did.

The equation r = mV/|q|B is the formula for the radius of curvature of a charged particle moving in a uniform magnetic field. In this case, the charged particle is a doubly charged helium atom with a mass of 6.68 x 10^-27 kg. The potential difference it is accelerated through is 4.00 x 10^3 V, and the magnetic field it is moving through is 0.450 T.

Plugging in the given values into the equation, we get:

r = (6.68 x 10^-27 kg)(4.00 x 10^3 V)/|2(1.6 x 10^-19 C)| (0.450 T)

= (2.672 x 10^-23 kg m^2 s^-2 C^-1)(4.00 x 10^3 V)/(3.2 x 10^-19 C) (0.450 T)

= (8.544 x 10^-20 kg m^2 s^-2 C^-1)(0.450 T)/(3.2 x 10^-19 C)

= 2.70 x 10^-19 m

So, the radius of curvature of the path of the atom in the given conditions is 2.70 x 10^-19 m.

I hope this explanation helps. Keep up the good work!
 

1. What is the radius of curvature of an atom's path?

The radius of curvature of an atom's path is the distance between the center of the atom and the center of its circular path. It is a measure of how much the atom's path curves.

2. How is the radius of curvature of an atom's path calculated?

The radius of curvature of an atom's path can be calculated using the equation R = mv/qB, where R is the radius of curvature, m is the mass of the atom, v is the velocity of the atom, q is the charge of the atom, and B is the strength of the magnetic field.

3. Does the radius of curvature change for different types of atoms?

Yes, the radius of curvature can vary for different types of atoms. It depends on the mass, charge, and velocity of the atom as well as the strength of the magnetic field.

4. How does the radius of curvature affect the behavior of atoms in a magnetic field?

The radius of curvature determines the path that an atom will take in a magnetic field. If the radius of curvature is small, the atom's path will be more curved, while a larger radius of curvature will result in a less curved path.

5. Can the radius of curvature of an atom's path be manipulated?

Yes, the radius of curvature can be manipulated by changing the strength of the magnetic field, the velocity of the atom, or the charge of the atom. This can be useful in controlling the behavior of atoms in various scientific experiments.

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