InterviewSolution
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InterviewSolution
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In a nuclear reactor, a neutron of high speed `(~~10^(7)ms^(-1))` must be slowed down to `10^(3)ms^(-1)` so that it can have a high probality of interacting with isotipe `_92U^(235)` and causing it to fission. Show that a neutron can lose most of its K.E. in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a fewe times the neutron mass. The material making up the light nuclei usually heavy water `(D_(2)O)` or graphite is called modertaor. |
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Answer» Initial K.E. of neutron is `K_(1)=(1)/(2)m_(1)u_(1)^(2)` Velocity of neutron after collision with deuterium, `upsilon_(1)=((m_(1)-m_(2))u_(1))/(m_(1)+m_(2))` Final K.E. of neutron is `K_(2)=(1)/(2)m_(1)upsilon_(1)^(2)=(1)/(2)m_(1)((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)u_(1)^(2)` `:.` Frational K.E. retained by neutron is `f_(1)=(K_(2))/(K_(1))=((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)` For deuterium, `m_(2)=2m_(1) :. f_(1)=(1)/(9)`, Fractional K.E. lost by neutron `f_(2)=1-f_(1)=1-(1)/(9)=(8)/(9)=90%` This is the fractional K.E. gained by moderating nuclei. Therefore, almost `90% ` of neutron energy is transferred to deuterium. Similarly, in case of carbon, we can show that `f_(1)=28.4%` and `f_(2)=71.6%`. |
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