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

Explanation:

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

             r = $\frac{mv}{qB}$

where,   m = mass,      v = velocity

              q = charge,    B = magnetic field

Therefore, q will be calculated as follows.

         q = $\frac{mv}{rB}$

            = $\frac{6.6 \times 10^{-27} \times 5.3 \times 10^{5}}{0.014 m \times 0.78 T}$

            = $\frac{34.98 \times 10^{-22}}{0.01092}$

            = $3.2 \times 10^{-19} \times \frac{1.0 e}{1.6 \times 10^{-19}C}$

            = +2e

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