### Related Formula Q = Delta M c^2$Q = \Delta M c^2$ Q = sum K.E._textproducts$Q = \sum K.E._{\text{products}}$ ### Core Logic The nuclear fission reaction can be written as: (mathrmX) rightarrow (mathrmY) + (mathrmZ) + (mathrmP)$(\mathrm{X}) \rightarrow (\mathrm{Y}) + (\mathrm{Z}) + (\mathrm{P})$ The parent mass is M$M$. Three similar daughter nuclei are formed, each with mass approx fracM3$\approx \frac{M}{3}$. The total energy released due to the mass defect Delta M$\Delta M$ is Delta M c^2$\Delta M c^2$. This energy is equally distributed among the three identical daughter nuclei as kinetic energy (assuming parent is at rest). ### Step 1: Equate Energy Delta M c^2 = frac12 left(fracM3right) V^2 + frac12 left(fracM3right) V^2 + frac12 left(fracM3right) V^2$\Delta M c^2 = \frac{1}{2} \left(\frac{M}{3}\right) V^2 + \frac{1}{2} \left(\frac{M}{3}\right) V^2 + \frac{1}{2} \left(\frac{M}{3}\right) V^2$ Delta M c^2 = 3 times frac12 left(fracM3right) V^2$\Delta M c^2 = 3 \times \frac{1}{2} \left(\frac{M}{3}\right) V^2$ Delta M c^2 = frac12 M V^2$\Delta M c^2 = \frac{1}{2} M V^2$ ### Step 2: Solve for V V^2 = frac2 Delta M c^2M$V^2 = \frac{2 \Delta M c^2}{M}$ V = c sqrtfrac2 Delta MM$V = c \sqrt{\frac{2 \Delta M}{M}}$ ### Pattern Recognition Since total mass of the products is M$M$ (ignoring the tiny mass defect for kinetic energy calculations), the total kinetic energy frac12 M V^2$\frac{1}{2} M V^2$ equals the released energy Delta M c^2$\Delta M c^2$. The number of identical fragments doesn't change the velocity expression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei