Question: A nuclear fission is represented by the following reaction \({U^{236}} = {X^{111}} + {Y^{122}} + 3...

Question

A nuclear fission is represented by the following reaction
${U^{236}} = {X^{111}} + {Y^{122}} + 3n$
If the binding energies per nucleon of ${X^{111}}$ , ${Y^{122}}$ and ${U^{236}}$ are $8.6MeV$ , $8.5MeV$ and $7.6MeV$ respectively, then the binding energy released in the reaction will be
A. $200MeV$
B. $202MeV$
C. $195MeV$
D. $198MeV$

Answer 1

Solution

To solve this question, we will utilise the extremely simple formula of energy released to determine the energy released in the following nuclear reaction, allowing us to rapidly obtain our desired response.

Formula used:
Energy released = $\sum {B.{E_P}} - B.{E_R}$
Here, $B.{E_P}$ is the binding energy of the product and $B.{E_R}$ is the binding energy of the reactant.

Complete step by step answer:
Let's begin by understanding the fundamental words relevant to this question, such as nuclear fission and binding energy.Nuclear fission is the splitting of an atom's nucleus into two or more smaller nuclei. Even by the intense standards of radioactive decay, the fission process frequently produces gamma photons and releases a considerable quantity of energy.

The amount of energy necessary to detach a particle from a system of particles or to disperse all of the particles in the system is referred to as binding energy. Subatomic particles in atomic nuclei, electrons bound to nuclei in atoms, and atoms and ions linked together in crystals are all examples of binding energy.

Now, coming to the given question; for binding energy there is a very basic formula to find out the energy released. Energy released is equal to sum of binding energy of product minus sum of binding energy of reactant i.e.
Energy released = $\sum {B.{E_P}} - B.{E_R}$
$\text{Energy released}= 111 \times 8.6 + 122 \times 8.5 - 236 \times 7.6 \\\ \Rightarrow \text{Energy released}= 954.0 + 1037.0 - 173.6 \\\ \therefore \text{Energy released}= 198\,MeV$
Thus, the energy released in the reaction will be $198\,MeV$ .

So, the correct option is D.

Note: The stability of a nucleus is essentially determined by the binding energy per nucleon. The higher the binding energy per nucleon, the more work is required to remove the nucleon from the nucleus, and hence the nucleus is more stable.