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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.

a
0.00352 meV
b
3.27 MeV
c
0.82 MeV
d
2.45 MeV

detailed solution

Correct option is B

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

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The daughter nuclei are unstable therefore they decay into stable end products M 4298o and X 54136e by successive emission of β-particles. Let mass of   01n=1.0087 amu, mass of  92235U=236.0526 amu, mass of  54136Xe=135.9170 amu, mass of M 4298o = 97.9054 


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