Nucleus A is having mass number 220 and its binding energy per nucleon is 5.6 MeV. It splits in two fragments ‘B’ and ‘C’ of mass numbers 104 and 116. The binding energy of nucleons in ‘B’ and ‘C’ is 6.4 MeV per nucleon. The energy Q released per fission will be:
Question
Nucleus A is having mass number 220 and its binding energy per nucleon is 5.6 MeV. It splits in two fragments ‘B’ and ‘C’ of mass numbers 104 and 116. The binding energy of nucleons in ‘B’ and ‘C’ is 6.4 MeV per nucleon. The energy Q released per fission will be:
Solution
To find the energy released per fission (Q), we can use the equation:
Q = (Δm)c^2
where Δm is the change in mass and c is the speed of light.
Step 1: Calculate the change in mass (Δm) for nucleus A: Δm = mass of nucleus A - (mass of nucleus B + mass of nucleus C) Δm = 220 - (104 + 116) Δm = 220 - 220 Δm = 0
Step 2: Calculate the energy released per fission (Q): Q = (Δm)c^2 Q = (0)(c^2) Q = 0
Therefore, the energy released per fission (Q) is 0.
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