Question:

A nuclear power plant produces energy by utilizing the process of nuclear fission. Consider a nucleus of Uranium-235 (U-235) that undergoes fission, breaking into two smaller nuclei, along with the release of two or three neutrons and a significant amount of energy. Given that the mass of a U-235 nucleus is approximately 3.90 × 10^-25 kg, and the speed of the neutrons released in the fission process is approximately 2.50 × 10^5 m/s, calculate the kinetic energy of a single neutron.

Answer:

To find the kinetic energy of a single neutron released during the fission process, we can use the equation for kinetic energy:

K.E. = (1/2)mv^2

Where:

• K.E. is the kinetic energy of the neutron,
• m is the mass of the neutron, and
• v is the speed of the neutron.

Given that the mass of a neutron is approximately 1.67 × 10^-27 kg and the speed of the neutron is 2.50 × 10^5 m/s, we can substitute these values into the formula and calculate the kinetic energy.

K.E. = (1/2) * (1.67 × 10^-27 kg) * (2.50 × 10^5 m/s)^2

Simplifying the expression:

K.E. = (1/2) * (1.67 × 10^-27 kg) * (6.25 × 10^10 m^2/s^2)

Multiplying the numbers:

K.E. = 1.67 × 10^-27 kg * 6.25 × 10^10 m^2/s^2

K.E. = 1.04 × 10^-16 J

Therefore, the kinetic energy of a single neutron released during the fission process is approximately 1.04 × 10^-16 Joules.