1.

In the fusion reaction ""_(1)H^(2)+ ""_(1)H^(2) tp ""_(2)He^(3) +""_(0)n^(1), douteron, helium and the neutron have masses 2.015 a.m.u., 3.017 a.mu, and 1-009 a.m.u. respectivoly. Find the total energy released if 1 kg of deuterium undergoes complete fusion :

Answer»

`3 xx 10^(12)`
`6 xx 10^(11)`
`9 xx 10^(13)`
`10.2 xx 10^(14)`

Solution :Mass difference `=2 xx 2.015 -(3.017+1.009)=0.004 a.m.u.`
Energy released `=0.004 xx 931MeV`
=3.724MeV
Energy released PER deuteron `=1/2 xx 3.724 MeV`
No. of deuterons in one KG `=(6.02 xx 10^(26))/(2)`
Energy per kg `=9 xx 10^(13)J`


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