Question

R9 Relativity and Modern Physics 46 o neutrons with total kinetic energy 7 MeV fuse into a Helium- R9.6.5.Two protons and two 4 nucl A. Setup with a labeled diagram. b. Write down the wit cus. The binding energy of the He-4 is 18.3 MeV nergy of the H e energy equation for the above process representing by M the mass of f He-4. From the above equation obtain the kinetic energy of the He nucleus in Mev. A. SETUP He ??- KE

Answer 1

A) The setup is shown in the problem

B) Binding energy of the Helium nucleus is

$18.3=2n+2p-M$

where n and p are the masses of neutrons and protons respectively

Therefore the mass of He nucleus is

$M=2n+2p-18.3$ ...(1)

The initial energy is masses of the neutrons and protons and 7 MeV of kientic energy

$2(n+p)+7$

This is equal to mass of He nucleus plus the kinetic energy

$2(n+p)+7=M+KE$ ...(2)

c) Therefore plugging in the value of M from (1) in euqation (2)

$2(n+p)+7=2(n+p)-18.3+KE$

Hence the final kinetic energy of the He nucleus is

$K=18.3+7=25.3\,\,MeV$