The binding energy of deuteron _1^2H is 1.112 | vedclass

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(C) The binding energy of a nucleus is given by the product of the number of nucleons and the binding energy per nucleon.For the reactant side,we have

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$23.8$

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(C) The binding energy of a nucleus is given by the product of the number of nucleons and the binding energy per nucleon.For the reactant side,we have two deuterons $(_{1}^{2}H)$:Total binding energy of two deuterons $= 2 	imes (2 	imes 1.112 \, MeV) = 4 	imes 1.112 \, MeV = 4.448 \, MeV$.For the product side,we have one alpha particle $(_{2}^{4}He)$:Total binding energy of one alpha particle $= 4 	imes 7.047 \, MeV = 28.188 \, MeV$.The energy released $Q$ in the fusion reaction is the difference between the total binding energy of the products and the total binding energy of the reactants:$Q = (28.188 \, MeV) - (4.448 \, MeV) = 23.74 \, MeV$.Rounding to the nearest provided option,we get $23.8 \, MeV$.

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

The binding energy of deuteron $_1^2H$ is $1.112 \, MeV$ per nucleon and an $\alpha$-particle $_2^4He$ has a binding energy of $7.047 \, MeV$ per nucleon. Then in the fusion reaction $_1^2H + _1^2H \to _2^4He + Q$,the energy $Q$ released is ........ $MeV$.

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