Question

What amount of energy is released in the fission of ${}_{92}\mathrm{U}^{235}$ ?

Answer 1

Fission:

• Nuclear fission is the process by which an atom's nucleus breaks into multiple smaller nuclei.
• Even by radioactive decay's energetic standards, the fission process frequently produces gamma photons and releases a considerable quantity of energy.

Fission of ${}_{92}\mathrm{U}^{235}$ :

$\mathrm{n}+{}_{92}{}^{235}\mathrm{U}\to {}_{36}{}^{92}\mathrm{Kr}+{}_{56}{}^{141}\mathrm{Ba}+3\mathrm{n}$

Step 1: analyzing the values

• $\mathrm{energy}\mathrm{released}=\mathrm{Q}=∆{\mathrm{mc}}^2$
• $1\mathrm{MeV}=1.609\mathrm{x}{10}^{-13}\mathrm{J}$
• $$\begin{gathered} {\mathrm{m}}_{\mathrm{n}}(\mathrm{mass}\mathrm{of}\mathrm{neutron})=1.00865\mathrm{u} \\ {\mathrm{m}}_{\mathrm{u}}(\mathrm{mass}\mathrm{of}\mathrm{uranium})=235.0439\mathrm{u} \\ {\mathrm{m}}_{\mathrm{k}}(\mathrm{mass}\mathrm{of}\mathrm{Kr})=91.92616\mathrm{u} \\ {\mathrm{m}}_{\mathrm{b}}(\mathrm{mass}\mathrm{of}\mathrm{barium})=140.9144\mathrm{u} \end{gathered}$$

Step 2:Finding mass defect:

• $$\begin{gathered} ∆\mathrm{m}=({\mathrm{m}}_{\mathrm{u}}+{\mathrm{m}}_{\mathrm{n}}-{\mathrm{m}}_{\mathrm{k}}-{\mathrm{m}}_{\mathrm{b}}-3\times {\mathrm{m}}_{\mathrm{n}}) \\ ∆\mathrm{m}=235+1-95-139-3\times 1 \\ ∆\mathrm{m}=0.1859\mathrm{u} \\ 1\mathrm{u}=1.66\times {10}^{-27}\mathrm{kg} \\ \mathrm{so}∆\mathrm{m}=3.09\times {10}^{-28}\mathrm{kg} \end{gathered}$$

Step 3: finding energy released

• $\mathrm{Q}=({\mathrm{m}}_{\mathrm{u}}+{\mathrm{m}}_{\mathrm{n}}-{\mathrm{m}}_{\mathrm{k}}-{\mathrm{m}}_{\mathrm{b}}-3\times {\mathrm{m}}_{\mathrm{n}}){(3\times {10}^9)}^2$
• $\mathrm{Q}=(3.09\times {10}^{-28}){(3\times {10}^9)}^2\mathrm{J}$
• $\mathrm{Q}=2.78\times {10}^{-11}\mathrm{J}$
• $$\begin{gathered} \mathrm{For}\mathrm{one}\mathrm{mole}\mathrm{of}{}_{92}{}^{235}\mathrm{U} \\ =2.78\times {10}^{-11}\times 6.022\times {10}^{23} \\ =1.67\times {10}^{10}\mathrm{kJ}{\mathrm{mol}}^{-1} \\ =1.67\times {10}^{13}\mathrm{J}{\mathrm{mol}}^{-1} \end{gathered}$$

So energy released in the fission of one mole of ${}_{92}\mathrm{U}^{235}$ is $1.67\times {10}^{13}\mathrm{J}{\mathrm{mol}}^{-1}$