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The nucleus ""_(92)^(235)Y , initially at rest, decays into ""_(90)^(213)X by emitting an alpha-particle ""_(92)^(235)Y to ""_(90)^(231)X + energy, The binding energies per nucleon of the parent nucleus, the daughter nucleus and alpha-particle are 7.8 MeV, 7.835 MeV and 7.07 MeV, respectively. Assuming the daughter nucleus to be formed in the unexcited state and neglecting its share in the energy of the reaction, find the speed of the emitted alpha-particle, ["Mass of particle "=6.68 xx 10^(-27) kg] |
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Answer» Solution :In the nuclear reaction `""_(92)^(235)Y to ""_(90)^(231)X+""_(2)^(4)He`+ energy, the energy released will be E=Total Binding energy of products-Total Binding energy of PARENT NUCLEUS `=(231 xx 7.835+4 xx 7.07-235 xx 7.8) MeV` =1809.885+28.280-183.0=5.165MeV `=5.165 xx 1.6 xx 10^(-13)J` This energy appearns as the KINETIC energy of alpha particle of mass `m=6.68 xx 10^(-27)KG`. Iv v be the speed of alpha particle, then `E=1/2 mv^(2)` `rArr v=sqrt((2E)/(m))= [(2 xx 5.165 xx 1.6 xx 10^(-13))/(6.68 xx 10^(-27))]^(1/2) =4.97 xx 10^(7)ms^(-1)` |
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