Question

What is the energy released in the fission reaction 1 0 n + 235 92 U → 131 53 I + 89 39 Y + 16 1 0 n 0 1 n + 92 235 U → 53 131 I + 39 89 Y + 16 0 1 n ? (The atomic mass of 131 I 131 I is 130.9061246 u and that of 89 Y 89 Y is 88.9058483 u)

Answer 1

Answer: The energy released in the given nuclear reaction is 94.99 MeV.

Explanation:

For the given nuclear reaction:

$_{92}^{235}\textrm{U}+_0^1\textrm{n}\rightarrow _{53}^{131}\textrm{I}+_{39}^{89}\textrm{Y}+16_{0}^1\textrm{n}$

We are given:

Mass of $_{92}^{235}\textrm{U}$ = 235.043924 u

Mass of $_{0}^{1}\textrm{n}$ = 1.008665 u

Mass of $_{53}^{131}\textrm{I}$ = 130.9061246 u

Mass of $_{39}^{89}\textrm{Y}$ = 88.9058483 u u

To calculate the mass defect, we use the equation:

$\Delta m=\text{Mass of reactants}-\text{Mass of products}$

Putting values in above equation, we get:

$\Delta m=(m_U+m_{n})-(m_{I}+m_{Y}+16m_{n})\\\\\Delta m=(235.043924+1.008665)-(130.9061246+88.9058483+16(1.008665))=0.1019761u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.1019761u)\times c^2$

$E=(0.1019761u)\times (931.5MeV)$  (Conversion factor:  $1u=931.5MeV/c^2$  )

$E=94.99MeV$

Hence, the energy released in the given nuclear reaction is 94.99 MeV.