Question

An ionized helium atom has a mass of 6.6 × 10-27 kg and a speed of 5.3 × 105 m/s. It moves perpendicular to a 0.78-T magnetic field on a circular path that has a 0.014-m radius. Determine whether the charge of the ionized atom is +e or +2e.

Answer 1

Step-by-step explanation:

In a magnetic field, the radius of the charged particle is as follows.

r =
`(mv)/(qB)`

where, m = mass, v = velocity

q = charge, B = magnetic field

Therefore, q will be calculated as follows.

q =
`(mv)/(rB)`

=
`(6.6 * 10^(-27) * 5.3 * 10^(5))/(0.014 m * 0.78 T)`

=
`(34.98 * 10^(-22))/(0.01092)`

=
`3.2 * 10^(-19) * (1.0 e)/(1.6 * 10^(-19)C)`

= +2e

Thus, we can conclude that the charge of the ionized atom is +2e.