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A deuterium reaction that occurs in an experimental fusion reactor is in two stapes \[\] (1) Two deuterium nucli fuse together to form a tritium nucleus , with a proton as a by-product written as $\; 1 > (D ,P)\; T\;.$ \[\] (2) A tritium nucleus fuses with another oleunterium nucleus to form a helium $\;_{2}He^{4}\;$ nucleus with neutron as a-by-product written as $\;T (D,n)\;_{2}He^{4}\;.$ Then the energy released in each of two stages and energy released in combined reaction is \[\] Given \[\] $\;_{1}D^{2}=2.014102\;a.m.u $ \[\] $\;_{1}T^{3}=3.016049\;a.m.u$ \[\] $\;_{1}H^{1}=1.007825\;a.m.u$ \[\] $\;_{1}n^{1}=1.008665\;a.m.u$

$(a)\;5.03 MeV\qquad(b)\;6.03 MeV\qquad(c)\;7.03 MeV\qquad(d)\;4.03 MeV$

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Answer : (d) $\;4.03 MeV$
Explanation :
The reaction is
$_{1}D^{2}+_{1}D^{2} \to 1_{T}^{3}+_{1}P^{1}+\varepsilon_{1}$
$\bigtriangleup m=[2\;(2.014)-3.016-1.0078]$
$=0.00433 a.m.u$
Energy released
$=4.03 MeV\;.$
answered Feb 24, 2014 by yamini.v

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