The energy released when 717.13 kg of 37Li is converted into 24He by proton bombardment is 1032 eV. The value of is . (Nearest integer) (Mass of 37Li=7.0183 u, mass of 24He=4.004 u, mass of proton =1.008 u and 1 u =931 MeV/c2 and Avogadro number =6.0 1023)
Answer
6
Solution
Reaction: 37Li+11H 2 24He
Mass defect: m=(7.0183+1.008)-2 4.004=8.0263-8.008=0.0183 u
Q=0.0183 931=17.04 MeV per reaction
Number of Li nuclei in 717.13 kg = 700017.13 g (molar mass 7 g):
N= 7000/17.137 6.0 1023= 100017.13 6.0 1023=3.50 1025
E=NQ=3.50 1025 17.04 106 eV
=5.97 1032 eV 6 1032 eV
∴ =6
Answer: 6