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 1385 V. What is the speed of the xenon ions as they exit the thruster?

Answer 1

To solve this problem we will apply the concept of energy conservation. For this specific case the potential electrical energy must be equivalent to the kinetic energy of the body. Therefore such equality will be reflected under the function,

Electric potential energy = Kinetic energy

Mathematically this is,

$qV = \frac{1}{2} mv^2$

Here,

q = Charge

V = Potential

m = mass

v = Velocity

Replacing with our values we have that,

$(1.6*10^{-19})(1385) = \frac{1}{2} (2.18*10^{-25})(v^2)$

Solving for v,

$v=45089.1m/s$

Therefore the speed of the xenon ions as they exit the thruster is 45089.1m/s