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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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The binding energy per nucleon of deuteron $\;(_{1}^{2}H)\;$ and helium nucleus $\;(_{2}^{4}He)\;$ is 1.1 MeV and 7 MeV respectively . If two deuteron nuclei react to form a single helium nucleus, the energy released is

$(a)\;23.6 MeV\qquad(b)\;22.8 MeV\qquad(c)\;19 MeV\qquad(d)\;23 MeV$

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Answer : $\;23.6 \;MeV$
If two deuteron nuclei react to form a single helium nucleus ,$^2_1 H + ^2_1 H \rightarrow ^4_2 H + Q$. We need to calculate Q.
Energy released = total binding energy of product - total binding energy of reactants
$\Rightarrow Q = 4\times 7 - 2 \times ( 2\times 1.1 ) = 28 - 4.4=23.6 \;MeV \;.$
answered Mar 6, 2014 by yamini.v
edited Aug 12, 2014 by balaji.thirumalai
 

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