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Sohail Kumar Bhagat
Asked: 2022-11-11T01:02:34+05:30 In:

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.

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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  1. 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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