Energy released in nuclear reactions -Q-value

Question

A certain mass of Hydrogen is changed to Helium by the process of fusion. The mass defect in fusion reaction is 0.02866 u. The energy liberated per u is (given 1 u = 931 MeV)

Easy

Solution

$As{}_{1}^{2}H{+}_{1}^{2}H{\to}_{2}^{4}He\phantom{\rule{0ex}{0ex}}Here,\Delta M=0.02866\text{\hspace{0.17em}}u\phantom{\rule{0ex}{0ex}}\therefore Theenergyliberatedperuis\phantom{\rule{0ex}{0ex}}=\frac{\Delta M\times 931}{4}MeV\phantom{\rule{0ex}{0ex}}=\frac{0.02866\times 931}{4}MeV=\frac{26.7}{4}MeV=6.675MeV$

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