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:
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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To solve this problem it is necessary to apply the related concepts to string vibration. This concept shows the fundamental frequency of a string due to speed and length, that is,
Where
v = Velocity
L = Length
Directly if the speed is maintained the frequency is inversely proportional to the Length:
Therefore the relationship between two frequencies can be described as
Our values are given as,
Therefore the second frequency is
The frequency allocation of 329Hz is note E.