CBSE Class 12-science Answered
The nucleus of an atom of , initally at rest decay by emitting an -particle as per the equation Energy. It is given that the binding energies per nucleon of the parent and the daughter nuclei are 7.8 MeV and 7.835 MeV respectively and that of - particle is 7.07 MeV/nucleon.Assuming the daughter nuleus to be formed in the unexcited state and neglecting the energy of the reaction, calculate the speed of the emitted - particle. Take mass of - particle to be 6.68 x 10-27 kg
Asked by Topperlearning User | 02 Jun, 2015, 09:08: AM
Expert Answer
Energy released, E = [ M( 231 X) + M(4He) - M(235Y)]c2
= [(7.835 x 231) + (7.07 x 4) - (7.8 x 235)]MeV
= [1809.9 + 28.28 - 1833]MeV
= 5.18MeV
= 5.18 x 1.6 x 10-13J
This entire kinetic energy is taken by - particle as given.
i.e., mv2 = 5.18 x 18.1.6 x 10-13J
x 6.68 x 10-27 v2=5.18 x 1.6 x 10-13J
Answered by | 02 Jun, 2015, 11:08: AM
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