The nuclear reaction for the proton bombardment of Lithium is:
37Li+11H→224He
The mass defect (Δm) for the reaction is:
Δm=m(37Li)+m(11H)−2m(24He)
Δm=7.0183+1.008−2(4.004)
Δm=8.0263−8.008=0.0183 u
Energy released per reaction (Q) is:
Q=Δm×931 MeV=0.0183×931 MeV=17.0373 MeV
Number of moles of 37Li in 17.137 kg:
n=7 g/mol17.137×103 g=17.131000 mol
Number of atoms of 37Li:
N=n×NA=17.131000×6.0×1023=17.136×1026
Total energy released (E) is:
E=N×Q=(17.136×1026)×(17.0373×106 eV)
E=17.13102.2238×1032 eV≈5.967×1032 eV
Rounding to the nearest integer, we get α=6.
Answer: 6