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The bombardment of Lithium with protons gives rise to the following reaction: `._3Li^7+._1H^1 to .2_He^4+.2_He^4+Q` The atomic masses of lithium, hydrogen and helium are : `7.016u, 1.008u` and `4.004u` resp. Find the initial energy of each of `alpha` particle. Take `1a.m.u=931MeV`. |
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Answer» Here, we are going atomic masses. In the given nuclear reaction, masses of 4 electrons before and after reaction cancel out. Therefore, we can use these masses as corresponding nuclear masses. `Q=Deltamxx931MeV` `=[m(Li)+m(H)-2m(He)]xx931MeV` `=[7.016+1.008-2xx4.004]xx931` `=0.016xx931=14.896MeV` This energy is shared equally between two alpha particles. `:.` Energy of each `alpha` particle `=14.896/2=7.448MeV` |
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