How many half-lives must go by for the radioactivity of this isotope to drop to (1/512) of its original value?
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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
Hence, the energy released in fission reaction is 191.715 MeV.
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