question_answer

Give reason.

(i) Lighter elements are better moderators for a nuclear reactor than heavier elements.

(ii) In a natural uranium reactor, heavy water is preferred moderator over ordinary water.

Or

How does the size of a nucleus depend on its mass number? Hence, explain why the density of nuclear matter of independent of the size of nucleus?

Answer:

(i) A moderator slows down fast neutrons released in a nuclear reactor. The basic principle of mechanics is that the energy transfer in a collision is the maximum when the colliding particles have equal masses. As, lighter elements have mass close to that of neutrons lighter elements are better moderators than heavier elements. (ii) Ordinary water has hydrogen nuclei \[(_{1}^{1}H)\] which have greater absorption capture for neutrons; so ordinary water will absorb neutrons rather than slowing them. On the other hand, the heavy hydrogen nuclei \[(_{1}^{2}H)\] have negligible absorption capture for neutrons, so they share energy with neutrons and neutrons are slowed down. Or The radius, R of nucleus is related to its mass number (A) as \[R={{R}_{0}}{{A}^{1/3}}\] where                \[{{R}_{0}}=1.1\times {{10}^{-15}}m\] If m is the average mass of a nucleon, then mass of nucleus = mA, where A is mass number. Volume of nucleus \[=\frac{4}{3}\pi {{R}^{3}}=\frac{4}{3}\pi {{({{R}_{0}}{{A}^{1/3}})}^{3}}=\frac{4}{3}\pi R_{0}^{3}A\] \[\therefore \] Density of nucleus, \[{{\rho }_{N}}=\frac{Mass}{Valume}\] or,        \[{{\rho }_{N}}=\frac{mA}{\frac{4}{3}\pi R_{0}^{3}A}=\frac{m}{\frac{4}{3}\pi R_{0}^{3}}=\frac{3m}{4\pi R_{0}^{3}}\] Clearly, nuclear density \[{{\rho }_{N}}\] is independent of mass number A.