Question

NASA's Deep Space 1 and Dawn spacecraft employed ion thrusters in which xenon ions with a charge of +1.60×10⁻¹⁹C and a mass of 2.18×10⁻²⁵kg are accelerated through a potential difference of 1480 V.

What is the speed of the xenon ions as they exit the thruster?

Answer 1

The speed of xenon ions exiting the thruster can be calculated using the work-energy principle, resulting in an approximation of 3.13×104 m/s, assuming no relativistic effects and that all kinetic energy comes from the electric field.

Step-by-step explanation:

The speed of the xenon ions as they exit the thruster can be calculated using the energy principle where the kinetic energy of the ions is equal to the work done on them by the electric field. The work done (W) is the product of the charge (q) of the ions and the potential difference (V) they are accelerated through, W = qV. The kinetic energy (KE) of an ion with mass (m) and speed (v) is given by KE = (1/2)mv2.

Equating the work done to the kinetic energy, we have qV = (1/2)mv2, so the speed v can be calculated by rearranging this equation to v = sqrt(2qV/m). Plugging in the values q = 1.60×10-19 C, V = 1480 V, and m = 2.18×10-25 kg, we can solve for v.

After performing the calculation we find the speed v of the xenon ions to be approximately 3.13×104 meters per second. It's important to note that this calculation is based on the assumption that the kinetic energy gained by the ions is solely due to the electric field and that relativistic effects are negligible.