Using the table below, find the Q values for the following reactions Atomic Mass
ID: 1393631 • Letter: U
Question
Using the table below, find the Q values for the following reactions Atomic Masses of the Neutron and Selected Isotopes Element Symbol Atomic mass, u Neutron Hydrogen 1.008 665 Protium 1.007 825 2.014 102 3.016 050 3.016 030 4.002 603 6.015 125 7.016 004 10.012 939 12.000 000 13.003 354 14.003 242 13.005 738 14.003 074 15.994 915 22.989 771 38.963 710 55.939 395 62.929 592 106.905 094 196.966 541 207.976 650 211.989 629 222.017 531 226.025 360 238.048 608 242.058 725 Deuterium H or D H or T 3He He Li Li Tritium Helium Lithium Boron Carbon 3C Nitrogen 14N Oxygen Sodium Potassium ron Copper Silver Gold Lead Polonium Radon Radium Uranium Plutonium 19 26 29 47 6 Cu 97 Au 208Pb 212Po 2Rn 226Ra 82 84 92 94 242 Mass values obtained at chttp://physics.nist.gov/PhysRefData/Compositions/index.html. (a) 1H +6Li 3He + 4He + Q Mev (b) 7Li + 1H 4He +4He + Q Mev (C) 2H + 3He 4He + 1H + Q MevExplanation / Answer
ANSWER:
In the reactions mass must me conserved. i.e. total mass before and after the reaction should be equal, hence fraction of mass converted into energy which is equal to Q.
a) In this reaction, before the reaction
Total mass of the reactants 1.007825 + 6.015125 = 7.02295 amu
Total mass of the products 3.016050 + 4.002603 = 7.018653 amu
Fractional mass lose or mass defect = 7.02295 - 7.018653 = 0.004297 amu = 0.004297 X 931.6 MeV
Q = 4.0030852 MeV
b) In this reaction, before the reaction
Total mass of the reactants 7.016004 + 1.007825 = 8.023829 amu
Total mass of the products 4.002603 + 4.002603 = 8.005206 amu
Fractional mass lose or mass defect = 0.018623 amu = 0.018623 X 931.6 MeV
Q = 17.3491868 MeV
c) In this reaction, before the reaction
Total mass of the reactants 2.014102 + 3.016030 = 5.020132 amu
Total mass of the products 4.002603 + 1.007825 = 5.010428 amu
Fractional mass lose or mass defect = 0.009704 amu = 0.009704 X 931.6 MeV
Q = 9.0402464 MeV
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