posted by Sarah .
An alpha particle, initially moving at a speed of 7.10x10^5m/s, goes through a potential difference of 2500V. Determine the final speed of the alpha particle.
Answer: 8.64 x 10^5
Multiply 2*e by the voltage change to get the additional kinetic energy.
deltaKE = 8.00*10^-16 J
initial KE = (Malpha)*Vo^2/2
The alpha particle has a mass of
= 4.00g/6.02*10^23 = 6.64*10^-24 g
= 6.64*10^-27 kg
Vo = 7.10^10^5 m/s
initial KE = 1.67*10^-15 J
Add initial KE and delta KE for the final KE.
final KE = 2.47*10^-15 J
Use the final kinetic energy to get the final velocity