Question

The deuteron, the nucleus of the deuterium atom ("heavy" hydrogen), consists of a proton and a neutron. It is observed experimentally than a high energy photon ("gamma ray") with a minimum energy of 2.2 MeVcan break up the deuteron into a free proton and a free neutron; this process is called "photodissociation." About what fraction of the deuteron rest energy corresponds to its Binding energy? The result shows that the deuteron is very lightly bound than the iron nucleus.

Short answer

The ratio of binding energy to rest energy for the deuteron is 0.118% .

Step-by-step solution

Identification of given Data

The given data is listed below,

The energy that is needed to break the deuteron is,$E_8=2.2\mathrm{MeV}$

Definition of Rest energy and Binding energy

The energy stored inside a stationary particle as a result of its mass is defined as the rest energy.

Binding energy, is the amount of energy required to separate a particle from a system of particles or disperse all the system’s particles.

Determination of rest energy of the deuteron

Write the expression for the rest energy of deuteron.

$E_{rest}=m_p{c}^2+m_n{c}^2+E_B$

Here,$m_p{c}^2$ is the rest energy of the proton that the value of 940 MeV ,$m_n{c}^2$ is the rest energy of the neutron that the value of 940 MeV, and$E_8$ is therest energy of the deuteron,which is the same as that of the energy that is needed to break the deuteron.

Substitute all the values in the above expression.

$$\begin{gathered} E_{rest}=940\mathrm{MeV}+940\mathrm{MeV}+2.2\mathrm{MeV} \\ =1882.2\mathrm{MeV} \end{gathered}$$

Determination of the ratio of binding energy to rest energy for the deuteron

Divide the deuteron’s binding energy by the rest energy of the deuteron.

$$\begin{gathered} \frac{E_B}{E_{rest.0_2}}=\frac{2.2}{1882.2} \\ =0.00188\times 100\% \\ =0.118\% \end{gathered}$$

Thus, the ratio of binding energy to rest energy for the deuteron is 0.118% .