- #1

Ante

- 3

- 0

## Homework Statement

*The deuterium nucleus starts out with a kinetic energy of 1.09e-13 joules, and the proton starts out with a kinetic energy of 2.19e-13 joules. The radius of a proton is 0.9e-15 m; assume that if the particles touch, the distance between their centers will be twice that.*

**A:**What will be the total kinetic energy of both particles an instant before they touch?**B:**What is the kinetic energy of the reaction products (helium nucleus plus photon)?

**C:**What was the gain of kinetic energy in this reaction? (The products have more kinetic energy than the original particles did when they were far apart. How much more?)

**D:**Kinetic energy can be used to drive motors and do other useful things. If a mole of hydrogen and a mole of deuterium underwent this fusion reaction, how much kinetic energy would be generated?

## Homework Equations

E_U= (1/(4(Pi)(Epsilon_0)))*(q_1*q_2)/r

E_F=E_i

## The Attempt at a Solution

A: Used the law of energy conservation

E=(1/(4(Pi)(Epsilon_0)))*(q_1*q_2)/r -(k_i)

B: Potential energy becomes kinetic energy: E=(1/(4(Pi)(Epsilon_0)))*(q_1*q_2)/r = K

C: (1/(4(Pi)(Epsilon_0)))*(q_1*q_2)/r - (k_i)

D:

One mole deuterium = 6.02*10^2p

One mol protons = 6.02*10^2p

One mole gives: 7.326*10^10 J