Chemical Kinetics and Nuclear ChemistryEasy
Question
13Al27 + 2He4 → 14Si30 + 1H1 + Q
Mass of 13Al27 = 26.9815 amu and mass of 14Si30 = 29.9738. The Q is equal to:
Mass of 13Al27 = 26.9815 amu and mass of 14Si30 = 29.9738. The Q is equal to:
Options
A.4.437 MeV
B.6.578 MeV
C.8.328MeV
D.2.329 MeV
Solution
Mass of 2He2 = 4.0026; 1H1 = 1.00783. (amu = 931 MeV)
Mass defect = (26.9815 + 4.0026) - (29.9738 + 1.0078)
= 30.9841 - 30.9816 = 0.0025 amu
= 0.0025 × 931 = 2.329MeV
Mass defect = (26.9815 + 4.0026) - (29.9738 + 1.0078)
= 30.9841 - 30.9816 = 0.0025 amu
= 0.0025 × 931 = 2.329MeV
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