An alpha particle (helium nucleus, charge +2e) starts from rest and travels a distance of 1.88 cm under the influence of a uniform electric field of magnitude 12.2 kV/m. What is the final kinetic energy of the alpha particle?
To find the final kinetic energy of the alpha particle, we can use the equation for work done by an electric field:
W = q * E * d
Where:
W is the work done by the electric field,
q is the charge of the particle,
E is the magnitude of the electric field,
and d is the distance traveled by the particle.
In this case, the charge of the alpha particle is +2e, where e is the elementary charge (1.6 x 10^-19 C). The electric field magnitude is given as 12.2 kV/m, which can be converted to volts per meter by multiplying by 1000. The distance traveled by the particle is 1.88 cm, which can be converted to meters by dividing by 100.
Let's substitute the values into the equation and solve for W:
W = (2 * 1.6 x 10^-19 C) * (12.2 x 1000 V/m) * (1.88 x 10^-2 m)
W = (3.2 x 10^-19 C) * (1.22 x 10^4 V/m) * (1.88 x 10^-2 m)
W = 7.624 x 10^-17 J
The work done by the electric field is equal to the change in kinetic energy of the particle. So, the final kinetic energy of the alpha particle is 7.624 x 10^-17 J.