The binding energy of deuteron $\ _{1}^{2}H$ is $1.112\ MeV$ per nucleon and an $\alpha$-particle $\ _{2}^{4}He$ has a binding energy of $7.047\ MeV$ per nucleon. Then in the fusion reaction $\ _{1}^{2}H + \ _{1}^{2}H \rightarrow_{2}^{4}He + Q$, the energy $Q$ released is

Question

Dr. Abubacker · Accepted Answer

(c) Energy equivalent to $\ _{1}H^{2} = 2 	imes 1.112 = 2.224\ MeV$
Energy equivalent to $\ _{2}{He}^{4} = 4 	imes 7.047 = 28.188\ MeV$
From the equation, energy released
$= 28.188 - 2 	imes 2.224 = 23.74\ MeV \approx 24\ MeV$

Q. The binding energy of deuteron $\ _{1}^{2}H$ is $1.112\ MeV$ per nucleon and an $\alpha$-particle $\ _{2}^{4}He$ has a binding energy of $7.047\ MeV$ per nucleon. Then in the fusion reaction $\ _{1}^{2}H + \ _{1}^{2}H \rightarrow_{2}^{4}He + Q$, the energy $Q$ released is

Explanation

Reviewed by Dr. Abubacker

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