A doubly charged helium atom (mass = 6.68 x 10-27 kg) is accelerated through a potential difference of 3.00x 103 V. What will be the radius of curvature of the path of the atom if it is in a uniform 0.410 T magnetic field?
The secod part of the question is...
What is the period of revolution for the atom in the previous question?
Get the velocity after acceleration in the electric field, then, with that velocity..
centripetal force= magnetic force
and solve for r.
To find the radius of curvature of the path of the atom in a magnetic field, we can use the formula:
r = (m * v) / (q * B)
where:
r = radius of curvature
m = mass of the atom
v = velocity of the atom
q = charge of the atom
B = magnetic field strength
First, let's calculate the charge of the helium atom. Since it is a doubly charged helium atom, its charge is 2 times the elementary charge, which is 1.60 x 10^-19 C. Therefore, q = 2 * 1.60 x 10^-19 C.
Next, let's calculate the velocity of the atom. We know the potential difference through which it is accelerated, so we can use the formula for the kinetic energy of a charged particle:
K.E. = q * V
where:
K.E. = kinetic energy
q = charge of the atom
V = potential difference
Since the atom has a charge of q = 2 * 1.60 x 10^-19 C and is accelerated through a potential difference of V = 3.00 x 10^3 V, we can calculate the kinetic energy. Plugging the values into the formula gives:
K.E. = (2 * 1.60 x 10^-19 C) * (3.00 x 10^3 V)
Now, we can use the formula for the kinetic energy of a moving object:
K.E. = (1/2) * m * v^2
Rearranging the formula, we can solve for v:
v = sqrt((2 * K.E.) / m)
Plugging in the values for K.E. and m, we can find v.
Finally, we can substitute the known values for m, v, q, and B into the formula for the radius of curvature:
r = (m * v) / (q * B)
Plugging in the values will give us the answer, which is the radius of curvature of the path of the atom in the magnetic field.