The deuterium nucleus starts out with a kinetic energy of 1.24 × 10-13 joules, and the proton starts out with a kinetic energy of 2.47 × 10-13 joules. The radius of a proton is 0.9 × 10-15 m; assume that if the particles touch, the distance between their centers will be twice that. What will be the total kinetic energy of both particles an instant before they touch?

The total kinetic energy of both particles is

Explanation:

Given that,

Kinetic energy of nucleus

Kinetic energy of proton

Radius of proton

We need to calculate the final potential energy

Using formula of final potential energy

Put the value into the formula

We need to calculate the initial energy of both the particles

Using formula of energy

We need to calculate the total kinetic energy of both particles

Using conservation of energy

Hence, The total kinetic energy of both particles is

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double the distance = divide strength by 4