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 105 and 115. The binding energy of nucleons in 'B' and 'C' is 6.4 MeV per nucleon. The energy Q released per fission will be :
Solution
<p><sup>220</sup>A $\to$ <sup>105</sup>B + <sup>115</sup>C</p>
<p>$\Rightarrow$ Q = [105 $\times$ 6.4 + 115 $\times$ 6.4] $-$ [220 $\times$ 5.6] MeV</p>
<p>$\Rightarrow$ Q = 176 MeV</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Nuclear Fission and Fusion
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