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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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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 $ \[\] $\;_{1}T^{3}=3.016049\;a.m.u$ \[\] $\;_{1}H^{1}=1.007825\;a.m.u$ \[\] $\;_{1}n^{1}=1.008665\;a.m.u$\[\] The energy released per deuterium nuclei is

$(a)\;29.3 MeV\qquad(b)\;24.4 MeV\qquad(c)\;21.6 MeV\qquad(d)\;18.6 MeV$

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Answer : (c) $\;21.6 MeV$
Explanation :
$_{1}T^{3}+_{1}D^{2} \to _{2}He^{4}+_{0}n^{1}+\varepsilon_{2}$
$\bigtriangleup m=[(3.016+2.014)-(4.0026+1.0086)]$
$=0.0188 a.m.u$
Therefore , energy released
$\varepsilon_{2}=0.01888 \times931 MeV=17.58 MeV$
Total energy released in both processess
$=17.58+4.03=21.611 MeV\;.$
answered Feb 24, 2014 by yamini.v
 

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