In the nuclear reaction H $$2$$ $$1$$ $$+H$$ $$2$$ $$1$$ →He $$3$$ $$2$$ $$+n$$ $$1$$ $$0$$ If the mass of the deuterium atom $$=$$ $$2.014741$$ anu, mass of He $$3$$ $$2$$ atom $$=$$ $$3.016977$$ amu and mass of neutron $$=$$ $$1.008987$$ amu, then the Q value of the reaction is nearly.

Question

In the nuclear reaction H $$2$$ $$1$$ $$+H$$ $$2$$ $$1$$ →He $$3$$ $$2$$ $$+n$$ $$1$$ $$0$$ If the mass of the deuterium atom $$=$$ $$2.014741$$ anu, mass of He $$3$$ $$2$$  atom $$=$$ $$3.016977$$ amu and mass of neutron $$=$$ $$1.008987$$ amu, then the Q value of the reaction is nearly.

Infinity Learn · Accepted Answer

$$Q=$$∑ $$Br-$$∑ $$BpC2Where$$ ∑ Br $$=$$ sum of the masses of reactantsand  ∑ $$Bp=sum$$ of the masses of the products∑ $$Br=2$$ x $$2.014741$$ amu∑ $$Bp=(3.016977)$$ amu∑ $$Bp=(3.016977+1.008987)$$ amu        $$=4.025964$$ amu∑ $$Br+$$∑ $$Bp=(4.029482-4.025964)$$ amu                          $$=0.003518$$ amuDecrease in mass appears as equivalent energy∴  Q $$=$$ $$0.003518$$ x $$931$$ MeV    $$=$$ $$3.27$$ MeV

In the nuclear reaction H 2 1 +H 2 1 →He 3 2 +n 1 0 If the mass of the deuterium atom = 2.014741 anu, mass of He 3 2 atom = 3.016977 amu and mass of neutron = 1.008987 amu, then the Q value of the reaction is nearly.

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