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Found in: Page 517

### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

# In both D-D reactions in equation (11-18). Two deuterons fuse to produce two particles, a nucleus of ${A}{=}{3}$ and a free nucleon. Mass decreases because the binding energy of the A=3 nucleus is greater than the combined binding energies of the two deuterons. The binding energy of helium -4 is even greater still. Why can't the deuterons simply fuse into a helium -4 nucleus and nothing else? Why must multiple particles be produced?

We need a multiple-particle process, so that the extra particle could carry away the excess energy.

See the step by step solution

## Step 1: Given data

Brass is an alloy of copper and zinc. Given that ${\rho }_{Brass}$ is higher than the ${\rho }_{\text{copper}}$ or ${\rho }_{zinc}$ at room temperature and drops much slower with a decrease in temperature.

## Step 2: Concept of Nuclear fission

Nuclear fission, subdivision of a heavy atomic nucleus, such as that of uranium or plutonium, is into two fragments of roughly equal mass.

${{\rho }}_{{\text{Brass}}}{>}{{\rho }}_{{\text{copper}}}{,}{{\rho }}_{zinc}$

## Step 3: Explanation of production of multiple particles in deuteron fusion despite simply

Suppose tile reaction of two deuterons going into single helium-4 did happen, and then choose a reference frame where the single helium-4 particle would be rest.

Initially, the energy equals to the mass energy of the two deuterons and their kinetic energies.

Finally, the energy is only the mass-energy of the helium-4 . Since the mass of two deuterons is bigger than the mass of the helium- 4 , the energy is not conserved.

Thus we need a multiple-particle process, so that the extra particle could carry away the excess energy.