Question

Find the energy released in the fission reaction¹₀n + ²³⁵₉₂U → ⁹⁸₄₀Zr + ¹³⁵₅₂Te +3(¹₀n) The atomic masses of the fission products are 97.912735 u for ⁹⁸₄₀Zr and 134.916450 u for ¹³⁵₅₂Te .

Answer 1

Fission reaction is given

¹₀n + ²³⁵₉₂U → ⁹⁸₄₀Zr + ¹³⁵₅₂Te +3(¹₀n)

The atomic masses of the fission products are 97.912735 u for ⁹⁸₄₀Zr and 134.916450 u for ¹³⁵₅₂Te. Hence, the energy released in fission reaction is 191.715 MeV.

How to find the energy released in the given fission reaction?

We know, that the atomic mass of the elements are as follows:

• ⁹⁸₄₀Zr - 97.9120u

• ¹³⁵₅₂Te - 134.9087u

• ²³⁵₉₂U  - 235.0483923u
• n = 1.008665u

In order to find the mass difference, we will calculate the initial mass and the mass of products.

The equation for the reaction:

n + ²³⁵₉₂U → ⁹⁸₄₀Zr + ¹³⁵₅₂Te +3n

Mass of initial reagents is 1.008665u + 235.0483923u = 233.052588u

Product of the reagents is 97.9120u + 134.9087u + 3(1.008665u) = 235.846773u

Now, using the formula of

$E=(\triangledown m)c^2$

$E=(0.205815u)\frac{931.494MeV/c^2}{u} c^2=191.715MeV$

Hence, the energy released in fission reaction is 191.715 MeV.

To learn more about fission reactor, refer to:

brainly.com/question/23276812

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