Skip to content
Skip to main content

About this free course

Author

Download this course

Share this free course

White dwarfs and neutron stars
White dwarfs and neutron stars

Start this free course now. Just create an account and sign in. Enrol and complete the course for a free statement of participation or digital badge if available.

5.1 Nuclear photodisintegration

There are many ways in which high-energy gamma rays can disintegrate nuclei of iron-56, but as an illustration of the physics involved, and the amount of energy that may be absorbed by this photodisintegration process, consider the situation in which a nucleus of iron-56 is broken down into helium nuclei:

gamma plus super 56 Fe long left right arrow 13 times super four He prefix plus of four n full stop

The amount of energy absorbed by this process in the forward direction is normal cap delta times cap q equals negative 124.4 MeV . At the temperature and density of a stellar core, the result is that about three-quarters of the iron nuclei are dissociated.

Activity 7

If each iron nucleus absorbs 124.4 MeV of energy by photodisintegration, and three-quarters of a core of mass 1.4 M is dissociated in this way, calculate the total energy absorbed by this process. (The mass of an iron-56 nucleus is 56u, where u equals 1.66 multiplication 10 super negative 27 kg is the atomic mass unit.)

Answer

The number of particles N in a sample of material is given by the total mass of the sample M divided by the individual masses of the particles m, i.e. N=M/m. Three-quarters of the core mass is cap m equals left parenthesis three solidus four right parenthesis multiplication 1.4 times cap m sub circled dot operator , so the number of iron-56 nuclei is

cap n equals left parenthesis three solidus four right parenthesis multiplication 1.4 times cap m sub circled dot operator solidus left parenthesis 56 times u right parenthesis
cap n equals 1.05 multiplication 1.99 multiplication 10 super 30 kg divided by 56 multiplication 1.66 multiplication 10 super negative 27 kg
cap n equals 2.25 multiplication 10 super 55 nuclei full stop

Each nucleus absorbs 124.4 MeV of energy, which is equivalent to 124.4 multiplication 10 super six eV prefix multiplication of 1.60 multiplication 10 super negative 19 cap j eV super negative one equals 1.99 multiplication 10 super negative 11 cap j . So the core absorbs 2.25 multiplication 10 super 55 multiplication 1.99 multiplication 10 super negative 11 cap j equals 4.5 multiplication 10 super 44 cap j via the photodisintegration of iron-56 nuclei.

At still higher temperatures, helium nuclei will also undergo photodisintegration as follows:

gamma plus super four He long left right arrow two p prefix plus of two n full stop

The energy absorbed by this process in the forward direction is normal cap delta times cap q equals negative 28.3 MeV . At the temperature and density of a stellar core, the result is that about half of the helium-4 nuclei are dissociated.

Activity 8

If each helium-4 nucleus absorbs 28.3 MeV of energy by photodisintegration, and half of the core of mass 1.4 M is dissociated in this way, calculate the total energy absorbed by this process. (The mass of a helium-4 nucleus is 4u, where u equals 1.66 multiplication 10 super negative 27 kg is the atomic mass unit.)

Comment

The number of particles is N = M/m. Half the core mass is cap m equals left parenthesis one solidus two right parenthesis multiplication 1.4 times cap m sub circled dot operator , so the number of helium-4 nuclei is

cap n equals left parenthesis one solidus two right parenthesis multiplication 1.4 times cap m sub circled dot operator solidus left parenthesis four times u right parenthesis
cap n equals 0.7 multiplication 1.99 multiplication 10 super 30 kg divided by four multiplication 1.66 multiplication 10 super negative 27 kg
cap n equals 2.10 multiplication 10 super 56 nuclei full stop

Each nucleus absorbs 28.3 MeV of energy, which is equivalent to 28.3 multiplication 10 super six eV prefix multiplication of 1.60 multiplication 10 super negative 19 cap j eV super negative one equals 4.53 multiplication 10 super negative 12 cap j full stop So the core absorbs 2.10 multiplication 10 super 56 multiplication 4.53 multiplication 10 super negative 12 cap j equals 9.5 multiplication 10 super 44 cap j via the photodisintegration of helium-4 nuclei.

Therefore, the total amount of energy absorbed by nuclear photodisintegration is 4.5 multiplication 10 super 44 cap j from the dissociation of iron-56 plus 9.5 multiplication 10 super 44 cap j from the dissociation of helium-4, or about 1.4 multiplication 10 super 45 cap j in total.